If `cos 2x + 2 cos x=1` then, `(2 - cos ^2x) sin^2x` is equal to

cos 2x = 2 cos^2 x - 1

If 5(〖tan〗^2⁡〖x-〖cos〗^2⁡x 〗 )=2 cos⁡〖2x+9,〗then the value of cos⁡4x is

If `cos2x+2cosx=1` then `sin^2x(2-cos^2x)` is

Prove that cos2x = cos^2x-sin^2x = 2cos^2x-1 = 1-2sin^2x = (1-tan^2x)/(1+tan^2x).

PreCalculus - 三角法: 三角恒等式 (31/57) cos^2(x)=(1+cos2x)/2 の証明

If `cos^(- 1)(2x^2-1)=2pi-2cos^(- 1)x` , then -

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lf `int ( sin 2x- cos 2 x) dx=1/(sqrt(2))sin (2x-a) + C`, then a=

Find values of sin x/2 cos x/2 and tan x/2 , Solving trigonometric questions ,@DhimanRajeshDhiman

Prove that sin^8 x−cos^8 x=(sin^2 x−cos^2 x)(1−2sin^2 xcos^2 x)

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if cos x + sin x equal root 2 cos x then what is the value of cot X

if ` sin x + sin^2 x = 1`, then the value of `cos^2 x + cos^4x` is

If cos x = (- sqrt(3))/2 and pi x (3pi)/2 then the value of 4cot^2 x - 3cos e * c ^ 2 * xis:

If `cos2x+2cosx=1t h e n ,(2-cos^2x)sin^2x` is equal to

`cos 2 theta =(sqrt(2)+1)(cos theta -(1)/(sqrt(2)))` の場合、`theta` の値は

x = a sin 2theta (1 + cos 2theta)、y = b cos 2theta (1 - cos 2theta) の場合、(dy)/(dx) = | 12 | 継続...