Consider a unit circle with centre O at the origin. Let A be the point (1,0).
Let P, Q and R be the points on the circle such that, arc AP = x
arc PQ = y
arc AR = −y
Then, arc AQ = arc AP + arc PQ = x+y.
Therefore, the coordinates of the points P, Q and R are,
P (cos x, sin x)
Q (cos(x+y), sin(x+y))
R (cos(−y), sin(−y))
We have arc PQ = arc RA
⇒ arc PQ + arc AP = arc RA + arc AP
⇒ arc AQ = arc RP
⇒ length of chord AQ = length of chord RP
[in a circle, equal arcs cut off equal chords]
⇒ AQ = RP ⇒ AQ$\displaystyle \small ^{2}$ = RP$\displaystyle \small ^{2}$
By using distance formula, [d$\displaystyle \small ^{2}$=(x$\displaystyle \small _{2}$ −x$\displaystyle \small _{1}$ )$\displaystyle \small ^{2}$+(y$\displaystyle \small _{2}$ −y$\displaystyle \small _{1}$ )$\displaystyle \small ^{2}$]
AQ$\displaystyle \small ^{2}$ = (cos(x+y)−1)$\displaystyle \small ^{2}$+(sin(x+y)−0)$\displaystyle \small ^{2}$
= cos$\displaystyle \small ^{2}$(x+y)+1−2cos(x+y)+sin$\displaystyle \small ^{2}$(x+y)
= cos$\displaystyle \small ^{2}$(x+y)+sin$\displaystyle \small ^{2}$(x+y)+1−2cos(x+y)
=1+1−2cos(x+y) [cos$\displaystyle \small ^{2}$x+sin$\displaystyle \small ^{2}$x=1]
AQ$\displaystyle \small ^{2}$ = 2−2cos(x+y)
RP$\displaystyle \small ^{2}$ = (cos x − cos(−y))$\displaystyle \small ^{2}$+(sin x−sin(−y))$\displaystyle \small ^{2}$
= (cos x − cos y)$\displaystyle \small ^{2}$+(sin x+ sin y)$\displaystyle \small ^{2}$ [cos(−y)=cos y & sin(−y)=−sin y]
= cos$\displaystyle \small ^{2}$x+cos$\displaystyle \small ^{2}$y−2cos x cos y+sin$\displaystyle \small ^{2}$x+sin$\displaystyle \small ^{2}$y+2sin x sin y
= (cos$\displaystyle \small ^{2}$x+sin$\displaystyle \small ^{2}$x)+(cos$\displaystyle \small ^{2}$y+sin$\displaystyle \small ^{2}$y)−2(cos x cos y−sin x sin y)
= 1+1− 2(cos x cos y−sin x sin y) [cos$\displaystyle \small ^{2}$x+sin$\displaystyle \small ^{2}$x=1]
RP$\displaystyle \small ^{2}$= 2− 2(cos x cos y−sin x sin y)
Since, AQ$\displaystyle \small ^{2}$ = RP$\displaystyle \small ^{2}$
2−2cos(x+y) = 2− 2(cos x cos y−sin x sin y)
−2cos(x+y) = − 2(cos x cos y−sin x sin y)
cos(x+y) = cos x cos y−sin x sin y ...(i)
Replacing y by −y in (i), we get
cox(x+(−y))= cos x cos (−y) − sin x sin(−y) [cos(−y)=cos y & sin(−y)=−sin y]
cos (x−y) = cos x cos y + sin x sin y ...(ii)
$\cos \left ( \frac{\pi }{2}-x \right )=\sin x$
$\cos \left ( \frac{\pi }{2}-x \right )=\cos \frac{\pi }{2}\cos x+\sin \frac{\pi }{2}\sin x$
= 0*cos x + 1*sin x = sin x
$\cos \left ( \frac{\pi }{2}+x \right )=-\sin x$
$\cos \left ( \frac{\pi }{2}+x \right )=\cos \frac{\pi }{2}\cos x-\sin \frac{\pi }{2}\sin x$
= 0*cos x − 1*sin x = −sin x
$\sin \left ( \frac{\pi }{2}-x \right )=\cos x$
$\sin \left ( \frac{\pi }{2}-x \right )=\cos \left [ \frac{\pi }{2}-\left ( \frac{\pi }{2}-x \right ) \right ]$ [$\cos \left ( \frac{\pi }{2}-x \right )=\sin x$]
= cos x
sin(x+y) = sin x cos y + cos x sin y
sin(x+y) = $\cos \left [ \frac{\pi }{2}-\left ( x+y \right ) \right ]$
= $\cos \left [ \left ( \frac{\pi }{2}-x \right )-y \right ]$
= $\cos \left ( \frac{\pi }{2}-x \right )\cos y+\sin \left ( \frac{\pi }{2}-x \right )\sin y$
sin(x+y) = sin x cos y + cos x sin y...(iii) [$\cos \left ( \frac{\pi }{2}-x \right )=\sin x$ and $\sin \left ( \frac{\pi }{2}-x \right )=\cos x$ ]
sin(x−y) = sin x cos y − cos x sin y
Replacing y by −y in (iii), we get
sin(x+(−y))=sin x cos (−y) + cos x sin (−y)
sin(x−y) = sin x cos y − cos x sin y [cos(−y)=cos y & sin(−y)=−sin y]
If none of x, y and (x+y) is an odd multiple of $\frac{\pi }{2}$ , then
$\tan \left ( x+y \right )=$ $\large \frac{\tan x+\tan y}{1-\tan x\tan y}$
Since, none of x, y and (x+y) is an odd multiple of $\frac{\pi }{2}$, it follows that none of cos x, cos y and cos(x+y) is zero.
$\tan \left ( x+y \right )=$ $\frac{\sin \left ( x+y \right )}{\cos \left ( x+y \right )}$
= $\large \frac{\sin x \cos y+\cos x\sin y}{\cos x \cos y-\sin x\sin y}$
Divide numerator and denominator by cos x cos y
= $\LARGE \frac{\frac{\sin x\cos y}{\cos x\cos y}+\frac{\cos x\sin y}{\cos x\cos y}}{\frac{\cos x\cos y}{\cos x\cos y}-\frac{\sin x\sin y}{\cos x\cos y}}$
= $\LARGE \frac{\frac{\sin x}{\cos x}+\frac{\sin y}{\cos y}}{1-\frac{\sin x}{\cos x}\frac{\sin y}{\cos y}}$
= $\large \frac{\tan x+\tan y}{1-\tan x\tan y}$
$\tan \left ( x-y \right )=$ $\large \frac{\tan x-\tan y}{1+\tan x\tan y}$
We know that,
$\tan \left ( x+y \right )=$ $\large \frac{\tan x+\tan y}{1-\tan x\tan y}$
Replace y by −y,
$\tan \left ( x+(-y) \right )=$ $\large \frac{\tan x+\tan (-y)}{1-\tan x\tan (-y)}$
$\tan \left ( x-y \right )=$ $\large \frac{\tan x-\tan y}{1+\tan x\tan y}$ [tan(−y)=−tan y]
If none of x, y and (x+y) is a multiple of Ļ, then
$\cot \left ( x+y \right )$ $\large =\frac{\cot x\cot y-1}{\cot y+\cot x}$
Since, none of x, y and (x+y) is a multiple of Ļ, it follows that none of sin x, sin y and sin(x+y) is zero.
$\cot \left ( x+y \right )$$\large =\frac{\cos\left ( x+y \right )}{\sin\left ( x+y \right )}$
=$\large \frac{\cos x\cos y-\sin x\sin y}{\sin x\cos y+\cos x\sin y}$
Divide numerator and denominator by sin x sin y,
= $\LARGE \frac{\frac{\cos x\cos y}{\sin x\sin y}-\frac{\sin x\sin y}{\sin x\sin y}}{\frac{\sin x\cos y}{\sin x\sin y}+\frac{\cos x\sin y}{\sin x\sin y}}$
= $\LARGE \frac{\frac{\cos x}{\sin x}\frac{\cos y}{\sin y}-1}{\frac{\cos y}{\sin y}+\frac{\cos x}{\sin x}}$
= $\large \frac{\cot x\cot y-1}{\cot y+\cot x}$
$\cot \left ( x-y \right )$ $\large =\frac{\cot x\cot y+1}{\cot y-\cot x}$
We know that,
cot(x+y) = $\large \frac{\cot x\cot y-1}{\cot y+\cot x}$
Replace y by −y,
cot(x+(−y)) = $\large \frac{\cot x\cot (-y)-1}{\cot (-y)+\cot x}$
cot(x+y) = $\large \frac{-\cot x\cot y-1}{-\cot y+\cot x}$ [cot(−y)=−cot y]
= $\large \frac{\cot x\cot y+1}{\cot y-\cot x}$
$\cos 2x$ $=\cos ^{2}x-\sin^{2}x$ $=2\cos^{2}x-1$ $=1-2\sin^{2}x$ $\large =\frac{1-\tan^{2}x}{1+\tan^{2}x}$
We know that,
cos(x+y) = cos x cos y − sin x sin y
Replacing y by x, we get
cos(x+x) = cos x cos x − sin x sin x
cos 2x = cos$\displaystyle \small ^{2}$x−sin$\displaystyle \small ^{2}$x
= cos$\displaystyle \small ^{2}$x−(1−cos$\displaystyle \small ^{2}$x)
=2cos$\displaystyle \small ^{2}$x−1
Again, cos 2x = cos$\displaystyle \small ^{2}$x−sin$\displaystyle \small ^{2}$x
= 1−sin$\displaystyle \small ^{2}$x −sin$\displaystyle \small ^{2}$x
= 1−2sin$\displaystyle \small ^{2}$x
Again, cos 2x = cos$\displaystyle \small ^{2}$x−sin$\displaystyle \small ^{2}$x
$\large =\frac{\cos^{2}x-\sin^{2}x}{1}$
$\large =\frac{\cos^{2}x-\sin^{2}x}{\cos^{2}x+\sin^{2}x}$
Divide numerator and denominator by cos$\displaystyle \small ^{2}$x,
=$\LARGE \frac{\frac{\cos^{2}x}{\cos^{2}x}-\frac{\sin^{2}x}{\cos^{2}x}}{\frac{\cos^{2}x}{\cos^{2}x}+\frac{\sin^{2}x}{\cos^{2}x}}$
=$\large \frac{1-\tan^{2}x}{1+\tan^{2}x}$
$ \sin 2x=2\sin x\cos x$ $\large =\frac{2\tan x}{1+\tan^{2}x}$
We know that,
sin(x+y) = sin x cos y+cos x sin y
Replacing y by x, we get
sin(x+x) = sin x cos x + cos x sin x
sin 2x= 2 sin x cos x
Again, sin 2x= 2 sin x cos x
$\large =\frac{2\sin x\cos x}{1}$
$\large =\frac{2\sin x\cos x}{\cos^{2}x+\sin^{2}x}$
Divide numerator and denominator by cos$\displaystyle \small ^{2}$x,
$\LARGE =\frac{\frac{2\sin x\cos x}{\cos^{2}x}}{\frac{\cos^{2}x}{\cos^{2}x}+\frac{\sin^{2}x}{\cos^{2}x}}$
$\large =\frac{2\tan x}{1+\tan^{2}x}$
$\tan 2x$ $\large =\frac{2\tan x}{1-\tan^{2}x}$
We know that,
$\tan \left ( x+y \right )=$ $\large \frac{\tan x+\tan y}{1-\tan x\tan y}$
Replacing y by x, we get
$\tan \left ( x+x \right )=$ $\large \frac{\tan x+\tan x}{1-\tan x\tan x}$
$\tan 2x$ $\large =\frac{2\tan x}{1-\tan^{2}x}$
sin 3x = 3sin x− 4 sin$\displaystyle \small ^{3}$x
sin 3x = sin(2x+x)
= sin 2x cos x + cos 2x sin x [sin(x+y)=sin x cos y+ cos x sin y]
= 2 sin x cos x cos x + (1−2sin$\displaystyle \small ^{2}$x) sin x [sin 2x=2sin x cos x & cos$\displaystyle \small ^{2}$x=1−2sin$\displaystyle \small ^{2}$x]
= 2 sin x cos$\displaystyle \small ^{2}$x + sin x − 2sin$\displaystyle \small ^{3}$x
= 2 sin x (1−sin$\displaystyle \small ^{2}$x) + sin x − 2sin$\displaystyle \small ^{3}$x [cos$\displaystyle \small ^{2}$x=1−sin$\displaystyle \small ^{2}$x]
= 2 sin x−2 sin$\displaystyle \small ^{3}$x + sin x − 2sin$\displaystyle \small ^{3}$x
= 3sinx−4sin$\displaystyle \small ^{3}$x
cos 3x = 4 cos$\displaystyle \small ^{3}$x−3cosx
cos 3x = cos(2x+x)
= cos 2x cos x − sin 2x sin x [cos(x+y)=cos x cos y−sin x sin y]
= (2cos$\displaystyle \small ^{2}$−1) cos x−2 sin x cos x sin x
= 2 cos$\displaystyle \small ^{3}$x−cos x − 2 sin$\displaystyle \small ^{2}$x cos x
= 2 cos$\displaystyle \small ^{3}$x−cos x− 2(1−cos$\displaystyle \small ^{2}$) cos x
= 2 cos$\displaystyle \small ^{3}$x−cos x − 2cos x + 2cos$\displaystyle \small ^{3}$x
= 4 cos$\displaystyle \small ^{3}$x−3cosx
$\cos x+\cos y$ $ =2\cos\left ( \frac{x+y}{2} \right )\cos\left ( \frac{x-y}{2} \right )$
$\cos x-\cos y$ $ =-2\sin\left ( \frac{x+y}{2} \right )\sin\left ( \frac{x-y}{2} \right )$
$\sin x+\sin y$ $ =2\sin\left ( \frac{x+y}{2} \right )\cos\left ( \frac{x-y}{2} \right )$
$\sin x-\sin y$ $ =2\cos\left ( \frac{x+y}{2} \right )\sin\left ( \frac{x-y}{2} \right )$
We know that,
cos(x+y) = cos x cos y − sin x sin y...(i)
cos(x−y) = cos x cos y + sin x sin y...(ii)
sin(x+y) = sin x cos y + cos x sin y...(iii)
sin(x−y) = sin x cos y − cos x sin y...(iv)
(i)+(ii) ⇒
cos(x+y) + cos(x−y) = 2 cos x cos y...(1)
(i)−(ii) ⇒
cos(x+y) − cos(x−y) = −2 sin x sin y...(2)
(iii)+(iv) ⇒
sin(x+y) + sin(x−y) = 2 sin x cos y...(3)
(iii)−(iv) ⇒
sin(x+y) − sin(x−y) = 2 cos x sin y...(4)
Let x+y = Īø and x−y = Ļ
⇒ $ x=\left ( \frac{\theta +\phi }{2} \right )$ and $ y=\left ( \frac{\theta -\phi }{2} \right )$
Substituting values of x and y in 1,2,3,4 ⇒
$\cos \theta +\cos \phi =2\cos\left ( \frac{\theta +\phi }{2} \right )\cos\left ( \frac{\theta -\phi }{2} \right )$
$\cos \theta -\cos \phi =-2\sin\left ( \frac{\theta +\phi }{2} \right )\sin\left ( \frac{\theta -\phi }{2} \right )$
$\sin\theta +\sin \phi =2\sin\left ( \frac{\theta +\phi }{2} \right )\cos\left ( \frac{\theta -\phi }{2} \right )$
$\sin\theta -\sin \phi =2\cos\left ( \frac{\theta +\phi }{2} \right )\sin\left ( \frac{\theta -\phi }{2} \right )$
Since Īø and Ļ can take any values, we can replace Īø by x and Ļ by y.
arc PQ = y
arc AR = −y
Then, arc AQ = arc AP + arc PQ = x+y.
Therefore, the coordinates of the points P, Q and R are,
P (cos x, sin x)
Q (cos(x+y), sin(x+y))
R (cos(−y), sin(−y))
We have arc PQ = arc RA
⇒ arc PQ + arc AP = arc RA + arc AP
⇒ arc AQ = arc RP
⇒ length of chord AQ = length of chord RP
[in a circle, equal arcs cut off equal chords]
⇒ AQ = RP ⇒ AQ$\displaystyle \small ^{2}$ = RP$\displaystyle \small ^{2}$
By using distance formula, [d$\displaystyle \small ^{2}$=(x$\displaystyle \small _{2}$ −x$\displaystyle \small _{1}$ )$\displaystyle \small ^{2}$+(y$\displaystyle \small _{2}$ −y$\displaystyle \small _{1}$ )$\displaystyle \small ^{2}$]
AQ$\displaystyle \small ^{2}$ = (cos(x+y)−1)$\displaystyle \small ^{2}$+(sin(x+y)−0)$\displaystyle \small ^{2}$
= cos$\displaystyle \small ^{2}$(x+y)+1−2cos(x+y)+sin$\displaystyle \small ^{2}$(x+y)
= cos$\displaystyle \small ^{2}$(x+y)+sin$\displaystyle \small ^{2}$(x+y)+1−2cos(x+y)
=1+1−2cos(x+y) [cos$\displaystyle \small ^{2}$x+sin$\displaystyle \small ^{2}$x=1]
AQ$\displaystyle \small ^{2}$ = 2−2cos(x+y)
RP$\displaystyle \small ^{2}$ = (cos x − cos(−y))$\displaystyle \small ^{2}$+(sin x−sin(−y))$\displaystyle \small ^{2}$
= (cos x − cos y)$\displaystyle \small ^{2}$+(sin x+ sin y)$\displaystyle \small ^{2}$ [cos(−y)=cos y & sin(−y)=−sin y]
= cos$\displaystyle \small ^{2}$x+cos$\displaystyle \small ^{2}$y−2cos x cos y+sin$\displaystyle \small ^{2}$x+sin$\displaystyle \small ^{2}$y+2sin x sin y
= (cos$\displaystyle \small ^{2}$x+sin$\displaystyle \small ^{2}$x)+(cos$\displaystyle \small ^{2}$y+sin$\displaystyle \small ^{2}$y)−2(cos x cos y−sin x sin y)
= 1+1− 2(cos x cos y−sin x sin y) [cos$\displaystyle \small ^{2}$x+sin$\displaystyle \small ^{2}$x=1]
RP$\displaystyle \small ^{2}$= 2− 2(cos x cos y−sin x sin y)
Since, AQ$\displaystyle \small ^{2}$ = RP$\displaystyle \small ^{2}$
2−2cos(x+y) = 2− 2(cos x cos y−sin x sin y)
−2cos(x+y) = − 2(cos x cos y−sin x sin y)
cos(x+y) = cos x cos y−sin x sin y ...(i)
Replacing y by −y in (i), we get
cox(x+(−y))= cos x cos (−y) − sin x sin(−y) [cos(−y)=cos y & sin(−y)=−sin y]
cos (x−y) = cos x cos y + sin x sin y ...(ii)
$\cos \left ( \frac{\pi }{2}-x \right )=\sin x$
$\cos \left ( \frac{\pi }{2}-x \right )=\cos \frac{\pi }{2}\cos x+\sin \frac{\pi }{2}\sin x$
= 0*cos x + 1*sin x = sin x
$\cos \left ( \frac{\pi }{2}+x \right )=-\sin x$
$\cos \left ( \frac{\pi }{2}+x \right )=\cos \frac{\pi }{2}\cos x-\sin \frac{\pi }{2}\sin x$
= 0*cos x − 1*sin x = −sin x
$\sin \left ( \frac{\pi }{2}-x \right )=\cos x$
$\sin \left ( \frac{\pi }{2}-x \right )=\cos \left [ \frac{\pi }{2}-\left ( \frac{\pi }{2}-x \right ) \right ]$ [$\cos \left ( \frac{\pi }{2}-x \right )=\sin x$]
= cos x
sin(x+y) = sin x cos y + cos x sin y
sin(x+y) = $\cos \left [ \frac{\pi }{2}-\left ( x+y \right ) \right ]$
= $\cos \left [ \left ( \frac{\pi }{2}-x \right )-y \right ]$
= $\cos \left ( \frac{\pi }{2}-x \right )\cos y+\sin \left ( \frac{\pi }{2}-x \right )\sin y$
sin(x+y) = sin x cos y + cos x sin y...(iii) [$\cos \left ( \frac{\pi }{2}-x \right )=\sin x$ and $\sin \left ( \frac{\pi }{2}-x \right )=\cos x$ ]
sin(x−y) = sin x cos y − cos x sin y
Replacing y by −y in (iii), we get
sin(x+(−y))=sin x cos (−y) + cos x sin (−y)
sin(x−y) = sin x cos y − cos x sin y [cos(−y)=cos y & sin(−y)=−sin y]
If none of x, y and (x+y) is an odd multiple of $\frac{\pi }{2}$ , then
$\tan \left ( x+y \right )=$ $\large \frac{\tan x+\tan y}{1-\tan x\tan y}$
Since, none of x, y and (x+y) is an odd multiple of $\frac{\pi }{2}$, it follows that none of cos x, cos y and cos(x+y) is zero.
$\tan \left ( x+y \right )=$ $\frac{\sin \left ( x+y \right )}{\cos \left ( x+y \right )}$
= $\large \frac{\sin x \cos y+\cos x\sin y}{\cos x \cos y-\sin x\sin y}$
Divide numerator and denominator by cos x cos y
= $\LARGE \frac{\frac{\sin x\cos y}{\cos x\cos y}+\frac{\cos x\sin y}{\cos x\cos y}}{\frac{\cos x\cos y}{\cos x\cos y}-\frac{\sin x\sin y}{\cos x\cos y}}$
= $\LARGE \frac{\frac{\sin x}{\cos x}+\frac{\sin y}{\cos y}}{1-\frac{\sin x}{\cos x}\frac{\sin y}{\cos y}}$
= $\large \frac{\tan x+\tan y}{1-\tan x\tan y}$
$\tan \left ( x-y \right )=$ $\large \frac{\tan x-\tan y}{1+\tan x\tan y}$
We know that,
$\tan \left ( x+y \right )=$ $\large \frac{\tan x+\tan y}{1-\tan x\tan y}$
Replace y by −y,
$\tan \left ( x+(-y) \right )=$ $\large \frac{\tan x+\tan (-y)}{1-\tan x\tan (-y)}$
$\tan \left ( x-y \right )=$ $\large \frac{\tan x-\tan y}{1+\tan x\tan y}$ [tan(−y)=−tan y]
If none of x, y and (x+y) is a multiple of Ļ, then
$\cot \left ( x+y \right )$ $\large =\frac{\cot x\cot y-1}{\cot y+\cot x}$
Since, none of x, y and (x+y) is a multiple of Ļ, it follows that none of sin x, sin y and sin(x+y) is zero.
$\cot \left ( x+y \right )$$\large =\frac{\cos\left ( x+y \right )}{\sin\left ( x+y \right )}$
=$\large \frac{\cos x\cos y-\sin x\sin y}{\sin x\cos y+\cos x\sin y}$
Divide numerator and denominator by sin x sin y,
= $\LARGE \frac{\frac{\cos x\cos y}{\sin x\sin y}-\frac{\sin x\sin y}{\sin x\sin y}}{\frac{\sin x\cos y}{\sin x\sin y}+\frac{\cos x\sin y}{\sin x\sin y}}$
= $\LARGE \frac{\frac{\cos x}{\sin x}\frac{\cos y}{\sin y}-1}{\frac{\cos y}{\sin y}+\frac{\cos x}{\sin x}}$
= $\large \frac{\cot x\cot y-1}{\cot y+\cot x}$
$\cot \left ( x-y \right )$ $\large =\frac{\cot x\cot y+1}{\cot y-\cot x}$
We know that,
cot(x+y) = $\large \frac{\cot x\cot y-1}{\cot y+\cot x}$
Replace y by −y,
cot(x+(−y)) = $\large \frac{\cot x\cot (-y)-1}{\cot (-y)+\cot x}$
cot(x+y) = $\large \frac{-\cot x\cot y-1}{-\cot y+\cot x}$ [cot(−y)=−cot y]
= $\large \frac{\cot x\cot y+1}{\cot y-\cot x}$
$\cos 2x$ $=\cos ^{2}x-\sin^{2}x$ $=2\cos^{2}x-1$ $=1-2\sin^{2}x$ $\large =\frac{1-\tan^{2}x}{1+\tan^{2}x}$
We know that,
cos(x+y) = cos x cos y − sin x sin y
Replacing y by x, we get
cos(x+x) = cos x cos x − sin x sin x
cos 2x = cos$\displaystyle \small ^{2}$x−sin$\displaystyle \small ^{2}$x
= cos$\displaystyle \small ^{2}$x−(1−cos$\displaystyle \small ^{2}$x)
=2cos$\displaystyle \small ^{2}$x−1
Again, cos 2x = cos$\displaystyle \small ^{2}$x−sin$\displaystyle \small ^{2}$x
= 1−sin$\displaystyle \small ^{2}$x −sin$\displaystyle \small ^{2}$x
= 1−2sin$\displaystyle \small ^{2}$x
Again, cos 2x = cos$\displaystyle \small ^{2}$x−sin$\displaystyle \small ^{2}$x
$\large =\frac{\cos^{2}x-\sin^{2}x}{1}$
$\large =\frac{\cos^{2}x-\sin^{2}x}{\cos^{2}x+\sin^{2}x}$
Divide numerator and denominator by cos$\displaystyle \small ^{2}$x,
=$\LARGE \frac{\frac{\cos^{2}x}{\cos^{2}x}-\frac{\sin^{2}x}{\cos^{2}x}}{\frac{\cos^{2}x}{\cos^{2}x}+\frac{\sin^{2}x}{\cos^{2}x}}$
=$\large \frac{1-\tan^{2}x}{1+\tan^{2}x}$
$ \sin 2x=2\sin x\cos x$ $\large =\frac{2\tan x}{1+\tan^{2}x}$
We know that,
sin(x+y) = sin x cos y+cos x sin y
Replacing y by x, we get
sin(x+x) = sin x cos x + cos x sin x
sin 2x= 2 sin x cos x
Again, sin 2x= 2 sin x cos x
$\large =\frac{2\sin x\cos x}{1}$
$\large =\frac{2\sin x\cos x}{\cos^{2}x+\sin^{2}x}$
Divide numerator and denominator by cos$\displaystyle \small ^{2}$x,
$\LARGE =\frac{\frac{2\sin x\cos x}{\cos^{2}x}}{\frac{\cos^{2}x}{\cos^{2}x}+\frac{\sin^{2}x}{\cos^{2}x}}$
$\large =\frac{2\tan x}{1+\tan^{2}x}$
$\tan 2x$ $\large =\frac{2\tan x}{1-\tan^{2}x}$
We know that,
$\tan \left ( x+y \right )=$ $\large \frac{\tan x+\tan y}{1-\tan x\tan y}$
Replacing y by x, we get
$\tan \left ( x+x \right )=$ $\large \frac{\tan x+\tan x}{1-\tan x\tan x}$
$\tan 2x$ $\large =\frac{2\tan x}{1-\tan^{2}x}$
sin 3x = 3sin x− 4 sin$\displaystyle \small ^{3}$x
sin 3x = sin(2x+x)
= sin 2x cos x + cos 2x sin x [sin(x+y)=sin x cos y+ cos x sin y]
= 2 sin x cos x cos x + (1−2sin$\displaystyle \small ^{2}$x) sin x [sin 2x=2sin x cos x & cos$\displaystyle \small ^{2}$x=1−2sin$\displaystyle \small ^{2}$x]
= 2 sin x cos$\displaystyle \small ^{2}$x + sin x − 2sin$\displaystyle \small ^{3}$x
= 2 sin x (1−sin$\displaystyle \small ^{2}$x) + sin x − 2sin$\displaystyle \small ^{3}$x [cos$\displaystyle \small ^{2}$x=1−sin$\displaystyle \small ^{2}$x]
= 2 sin x−2 sin$\displaystyle \small ^{3}$x + sin x − 2sin$\displaystyle \small ^{3}$x
= 3sinx−4sin$\displaystyle \small ^{3}$x
cos 3x = 4 cos$\displaystyle \small ^{3}$x−3cosx
cos 3x = cos(2x+x)
= cos 2x cos x − sin 2x sin x [cos(x+y)=cos x cos y−sin x sin y]
= (2cos$\displaystyle \small ^{2}$−1) cos x−2 sin x cos x sin x
= 2 cos$\displaystyle \small ^{3}$x−cos x − 2 sin$\displaystyle \small ^{2}$x cos x
= 2 cos$\displaystyle \small ^{3}$x−cos x− 2(1−cos$\displaystyle \small ^{2}$) cos x
= 2 cos$\displaystyle \small ^{3}$x−cos x − 2cos x + 2cos$\displaystyle \small ^{3}$x
= 4 cos$\displaystyle \small ^{3}$x−3cosx
$\cos x+\cos y$ $ =2\cos\left ( \frac{x+y}{2} \right )\cos\left ( \frac{x-y}{2} \right )$
$\cos x-\cos y$ $ =-2\sin\left ( \frac{x+y}{2} \right )\sin\left ( \frac{x-y}{2} \right )$
$\sin x+\sin y$ $ =2\sin\left ( \frac{x+y}{2} \right )\cos\left ( \frac{x-y}{2} \right )$
$\sin x-\sin y$ $ =2\cos\left ( \frac{x+y}{2} \right )\sin\left ( \frac{x-y}{2} \right )$
We know that,
cos(x+y) = cos x cos y − sin x sin y...(i)
cos(x−y) = cos x cos y + sin x sin y...(ii)
sin(x+y) = sin x cos y + cos x sin y...(iii)
sin(x−y) = sin x cos y − cos x sin y...(iv)
(i)+(ii) ⇒
cos(x+y) + cos(x−y) = 2 cos x cos y...(1)
(i)−(ii) ⇒
cos(x+y) − cos(x−y) = −2 sin x sin y...(2)
(iii)+(iv) ⇒
sin(x+y) + sin(x−y) = 2 sin x cos y...(3)
(iii)−(iv) ⇒
sin(x+y) − sin(x−y) = 2 cos x sin y...(4)
Let x+y = Īø and x−y = Ļ
⇒ $ x=\left ( \frac{\theta +\phi }{2} \right )$ and $ y=\left ( \frac{\theta -\phi }{2} \right )$
Substituting values of x and y in 1,2,3,4 ⇒
$\cos \theta +\cos \phi =2\cos\left ( \frac{\theta +\phi }{2} \right )\cos\left ( \frac{\theta -\phi }{2} \right )$
$\cos \theta -\cos \phi =-2\sin\left ( \frac{\theta +\phi }{2} \right )\sin\left ( \frac{\theta -\phi }{2} \right )$
$\sin\theta +\sin \phi =2\sin\left ( \frac{\theta +\phi }{2} \right )\cos\left ( \frac{\theta -\phi }{2} \right )$
$\sin\theta -\sin \phi =2\cos\left ( \frac{\theta +\phi }{2} \right )\sin\left ( \frac{\theta -\phi }{2} \right )$
Since Īø and Ļ can take any values, we can replace Īø by x and Ļ by y.
0 Comments