Advertisements
Advertisements
प्रश्न
Prove that: \[\cos 4x - \cos 4\alpha = 8 \left( \cos x - \cos \alpha \right) \left( \cos x + \cos \alpha \right) \left( \cos x - \sin \alpha \right) \left( \cos x + \sin \alpha \right)\]
Advertisements
उत्तर
\[RHS = 8\left( \text{ cos } x - cos \alpha \right) \left( \text{ cos } x + cos\alpha \right) \left( \text{ cos } x - sin\alpha \right) \left( \text{ cos } x + sin\alpha \right)\]
\[ = 8\left( \cos^2 x - \cos^2 \alpha \right) \left( \cos^2 x - \sin^2 \alpha \right)\]
\[ = 8\left( \cos^4 x - \cos^2 x \times \sin^2 \alpha - \cos^2 \alpha \times \cos^2 x + \cos^2 \alpha \times \sin^2 \alpha \right)\]
\[ = 8\left\{ \cos^4 x - \cos^2 x\left( \sin^2 \alpha + \cos^2 \alpha \right) + \cos^2 \alpha \times \sin^2 \alpha \right\}\]
\[ = 8\left\{ \cos^4 x - \cos^2 x + \cos^2 \alpha \times \left( 1 - \cos^2 \alpha \right) \right\}\]
\[ = 8\left\{ \cos^4 x - \cos^2 x + \cos^2 \alpha - \cos^4 \alpha \right\}\]
\[ = 8\left\{ \cos^2 x\left( \cos^2 x - 1 \right) + \cos^2 \alpha \times \left( 1 - \cos^2 \alpha \right) \right\}\]
\[= 8\left\{ \frac{1}{2} \cos^2 x\left( 2 \cos^2 x - 2 \right) + \frac{1}{2} \cos^2 \alpha \times \left( 2 - 2 \cos^2 \alpha \right) \right\}\]
\[ = 8\left\{ \frac{1}{2} \cos^2 x\left( 2 \cos^2 x - 1 - 1 \right) - \frac{1}{2} \cos^2 \alpha \times \left( 2 \cos^2 \alpha - 1 - 1 \right) \right\}\]
\[ = 8\left\{ \frac{1}{2} \cos^2 x\left( \cos2x - 1 \right) - \frac{1}{2} \cos^2 \alpha \times \left( \cos2\alpha - 1 \right) \right\} \left( \because \cos2\alpha = 2 \cos^2 \alpha - 1 \right) \]
\[ = 8\left[ \frac{1}{4}\left\{ 2 \cos^2 x\left( \cos2x - 1 \right) - 2 \cos^2 \alpha \times \left( \cos2\alpha - 1 \right) \right\} \right]\]
\[ = 8\left[ \frac{1}{4}\left\{ \left( 1 + \cos2x \right)\left( \cos2x - 1 \right) - \left( 1 + \cos2\alpha \right)\left( \cos2\alpha - 1 \right) \right\} \right]\]
\[= 8\left[ \frac{1}{4}\left\{ \cos^2 2x - 1 - \cos^2 2\alpha + 1 \right\} \right]\]
\[ = 8\left[ \frac{1}{8}\left\{ 2 \cos^2 2x - 2 \cos^2 2\alpha \right\} \right]\]
\[ = \left[ \left\{ \left( 1 + \cos4x \right) - \left( 1 + \cos4\alpha \right) \right\} \right] \]
\[ = \left[ 1 + \cos4x - 1 - \cos4\alpha \right]\]
\[ = \cos4x - \cos4\alpha = LHS\]
\[\text{ Hence proved } .\]
APPEARS IN
संबंधित प्रश्न
Prove that: \[\frac{\sin 2x}{1 + \cos 2x} = \tan x\]
Prove that: \[\frac{1 - \cos 2x + \sin 2x}{1 + \cos 2x + \sin 2x} = \tan x\]
Prove that: \[\frac{\cos x}{1 - \sin x} = \tan \left( \frac{\pi}{4} + \frac{x}{2} \right)\]
Prove that: \[\cos^2 \frac{\pi}{8} + \cos^2 \frac{3\pi}{8} + \cos^2 \frac{5\pi}{8} + \cos^2 \frac{7\pi}{8} = 2\]
Prove that: \[\sin^2 \left( \frac{\pi}{8} + \frac{x}{2} \right) - \sin^2 \left( \frac{\pi}{8} - \frac{x}{2} \right) = \frac{1}{\sqrt{2}} \sin x\]
Prove that: \[\cos 4x = 1 - 8 \cos^2 x + 8 \cos^4 x\]
Show that: \[2 \left( \sin^6 x + \cos^6 x \right) - 3 \left( \sin^4 x + \cos^4 x \right) + 1 = 0\]
If \[\cos x = - \frac{3}{5}\] and x lies in the IIIrd quadrant, find the values of \[\cos\frac{x}{2}, \sin\frac{x}{2}, \sin 2x\] .
If \[\cos x = \frac{4}{5}\] and x is acute, find tan 2x
If \[\sin x = \frac{4}{5}\] and \[0 < x < \frac{\pi}{2}\]
, find the value of sin 4x.
Prove that: \[\cos\frac{2\pi}{15} \cos\frac{4\pi}{15} \cos \frac{8\pi}{15} \cos \frac{16\pi}{15} = \frac{1}{16}\]
If \[2 \tan \alpha = 3 \tan \beta,\] prove that \[\tan \left( \alpha - \beta \right) = \frac{\sin 2\beta}{5 - \cos 2\beta}\] .
If \[\cos x = \frac{\cos \alpha + \cos \beta}{1 + \cos \alpha \cos \beta}\] , prove that \[\tan\frac{x}{2} = \pm \tan\frac{\alpha}{2}\tan\frac{\beta}{2}\]
If \[\cos \alpha + \cos \beta = \frac{1}{3}\] and sin \[\sin\alpha + \sin \beta = \frac{1}{4}\] , prove that \[\cos\frac{\alpha - \beta}{2} = \pm \frac{5}{24}\]
If \[\cos\alpha + \cos\beta = 0 = \sin\alpha + \sin\beta\] , then prove that \[\cos2\alpha + \cos2\beta = - 2\cos\left( \alpha + \beta \right)\] .
Prove that: \[\sin^2 24°- \sin^2 6° = \frac{\sqrt{5} - 1}{8}\]
Prove that: \[\cos 78° \cos 42° \cos 36° = \frac{1}{8}\]
If \[\cos 4x = 1 + k \sin^2 x \cos^2 x\] , then write the value of k.
If \[\frac{\pi}{2} < x < \frac{3\pi}{2}\] , then write the value of \[\sqrt{\frac{1 + \cos 2x}{2}}\]
Write the value of \[\cos^2 76° + \cos^2 16° - \cos 76° \cos 16°\]
If \[\frac{\pi}{4} < x < \frac{\pi}{2}\], then write the value of \[\sqrt{1 - \sin 2x}\] .
The value of \[\left( \cot \frac{x}{2} - \tan \frac{x}{2} \right)^2 \left( 1 - 2 \tan x \cot 2 x \right)\] is
The value of \[\tan x \sin \left( \frac{\pi}{2} + x \right) \cos \left( \frac{\pi}{2} - x \right)\]
\[\frac{\sin 3x}{1 + 2 \cos 2x}\] is equal to
The value of \[2 \sin^2 B + 4 \cos \left( A + B \right) \sin A \sin B + \cos 2 \left( A + B \right)\] is
If \[\tan \frac{x}{2} = \frac{\sqrt{1 - e}}{1 + e} \tan \frac{\alpha}{2}\] , then \[\cos \alpha =\]
If \[\left( 2^n + 1 \right) x = \pi,\] then \[2^n \cos x \cos 2x \cos 2^2 x . . . \cos 2^{n - 1} x = 1\]
If \[n = 1, 2, 3, . . . , \text{ then } \cos \alpha \cos 2 \alpha \cos 4 \alpha . . . \cos 2^{n - 1} \alpha\] is equal to
The greatest value of sin x cos x is ______.
If acos2θ + bsin2θ = c has α and β as its roots, then prove that tanα + tanβ = `(2b)/(a + c)`.
`["Hint: Use the identities" cos2theta = (1 - tan^2theta)/(1 + tan^2theta) "and" sin2theta = (2tantheta)/(1 + tan^2theta)]`.
If θ lies in the first quadrant and cosθ = `8/17`, then find the value of cos(30° + θ) + cos(45° – θ) + cos(120° – θ).
The value of `sin pi/10 sin (13pi)/10` is ______.
`["Hint: Use" sin18^circ = (sqrt5 - 1)/4 "and" cos36^circ = (sqrt5 + 1)/4]`
The value of `sin pi/18 + sin pi/9 + sin (2pi)/9 + sin (5pi)/18` is given by ______.
