Trigonometric Functions

31. L.H.S. = sin 2x + 2 sin 4x + sin 6x

Using sin A + sin B = 2 sin A + B/2 cos A - B/2  we have,

L.H.S. = (sin 2x + sin 6x) + 2 sin 4x

$=2sin\left(\frac{2x+6x}{2}\right)cos\left(\frac{2x-6x}{2}\right)+2sin4x.$

$=2sin\frac{8x}{2}cos\left(-\frac{4x}{2}\right)+2sin4x.$

$=2sin4xcos2x+2sin4x[?cos(-x)=x]$

$=2sin4x[cos2x+1]$

We know that,

$cos2x=2co{s}^{2}x-1$

$\Rightarrow cos2x+1=2co{s}^{2}x$

Hence,

L.H.S$=2sin4x\left(2co{s}^{2}x\right)$

$=4co{s}^{2}xsin4x$

= R.H.S

30. L.H.S $L.H.S=co{s}^{2}2x-co{s}^{2}6x$

$=(cos2x+cos6x)(cos2x-cos6x)\left[?a^2-b^2=(a-b)(a+b)\right]$

L.H.S = $L.H.S=\left[2cos\left(\frac{2x+6x}{2}\right)cos\left(\frac{2x-6x}{2}\right)\right]\left[-2sin\frac{2x+6x}{2}sin\left(\frac{2x-6x}{2}\right)\right]$

$=\left[2cos\frac{8x}{2}cos\left(-\frac{4x}{2}\right)\right]\left[-2sin\left(\frac{8x}{2}\right)sin\left(-\frac{4x}{2}\right)\right]$

$=2cos4xcos2x*2sin4x·sin2x[?cos(-x)=cosx;sin(-x)=-sinx]$

$=2sin4xcos4x*2sin2xcos2x$

As sin 2θ = 2 sinθ cosθ.

L.H.S. = sin (2*4x) sin(2*2x)

= sin 8x sin 4x

= R.H.S.

29. L.H.S  $L.H.S=si{n}^{2}6x-si{n}^{2}4x.$

$=(sin6x+sin4x)(sin6x-sin4x).\left[a^2-b^2=(a-b)(a+b)\right]$

So, L.H.S $L.H.S.=\left(2sin\left(\frac{6x+4x}{2}\right)cos\left(\frac{6x-4x}{2}\right)\right)\left(2cos\left(\frac{6x+4x}{2}\right)sin\left(\frac{6x-4x}{2}\right)\right)$

$=\left(2sin\frac{10x}{2}cos\frac{2x}{2}\right)\left(2cos\frac{10x}{2}sin\frac{2x}{2}\right)$

$=2sin5xcosx\cdot 2cos5xsinx$

$=2sin5xcos5x\cdot 2csinxcosx.$

As $sin2\theta =2sin\theta cos\theta$ we can write.

$L.H.S =sin\left(2*5x\right)=sin(2*x)$

$=sin10xsin2x= R.H.S.$ R.H.S

$\mathrm{28.}$$$$$ L.H.S = $L.H.S =cos\left(\frac{3\pi }{4}+x\right)-cos\left(\frac{3\pi }{4}-x\right)$

Using cos (A + B) = cos A cos B – sin A sin B

and cos (A – B) = cos A cos B + sin A sin B

$L.H.S =\left[cos\frac{3\pi }{4}cosx-sin\frac{3\pi }{4}sinx\right]-\left[cos\frac{3\pi }{4}cosx+sin\frac{3\pi }{4}sinx\right]$

$=cos\frac{3\pi }{4}cosx-sin\frac{3\pi }{4}sinx-cos\frac{3\pi }{4}cosx-sin\frac{3\pi }{4}sinx.$

$=-2sin\frac{3\pi }{4}sinx.$

$=-2sin\left(\frac{4\pi -\pi }{4}\right)sinx.$

27. L.H.S$L.HS=sin(x+1)xsin(x+2)x+cos(x+1)xcos(x+2)x.$

$=cos(x+1)xcos(x+2)x+sin(x+1)xsin(x+2)x.$

Let A = (n+1)x and B = (n+2)x

So, L.H.S = cosAcosB + sin Asin B

$=cos(A-B)$

Putting values of A and B we get,

L.H.S = $L.H.S = cos[(n+1)x-(n+2)x]$

$=cos[nx+x-nx-2x]$

$=cos(-x)$

$=cosx= R.H.S$ R.H.S

26. $$L.H.S $L.H.S.=cos\left(\frac{3\pi }{2}+x\right)cos(2\pi +x)\left[cot\left(\frac{3\pi }{2}-x\right)+cot(2\pi +x)\right]$

Here,

$cos\left(\frac{3\pi }{2}+x\right)=cos\left(\frac{4\pi -\pi }{2}+x\right)=cos\left(2\pi -\frac{\pi }{2}+x\right)$

$=cos\left[2\pi -\left(\frac{\pi }{2}-x\right)\right];VI quadrant.$

$cos\left(\frac{\pi }{2}-x\right);i I^{st} quadrant.$

$=sinx$

$cos(2x+x)=cosx,as xlies in I^{st}quadrant.$

$cot(2\pi +x)=cotx;asxliesin{I}^{st}quadrant.$

$cot\left(\frac{3\pi }{2}-x\right)=cot\left[\frac{4\pi -\pi }{2}-x\right]$

$=cot\left[2\pi -\frac{\pi }{2}-x\right]$

$=cot\left[2x-\left(\frac{\pi }{2}+x\right)\right];V{I}^{th} quadrant.$

$=-cot\left(\frac{\pi }{2}+x\right);I{I}^{nd}quadrat.$

$=-(-tanx)$

$=tanx.$

So.L.H.S $L.H.S=sinxcosx[tanx+cotx]$

$=sinxcosx\left[\frac{sinx}{cosx}+\frac{cosx}{sinx}\right]$

$=sinxcosx\frac{\left(si{n}^{2}x+co{s}^{2}x\right)}{cosxsinx}$

$=si{n}^{2}x+co{s}^{2}x$

= 1

= R.H.S.

23. (i) sin 75°= sin (45°+30°)

Using sin (x + y)= sin x cos y + cos x sin y we can write