Advertisements
Advertisements
प्रश्न
Prove that: \[\cos 4x - \cos 4\alpha = 8 \left( \cos x - \cos \alpha \right) \left( \cos x + \cos \alpha \right) \left( \cos x - \sin \alpha \right) \left( \cos x + \sin \alpha \right)\]
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: \[\frac{\sin x + \sin 2x}{1 + \cos x + \cos 2x} = \tan x\]
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: \[\sin^2 \frac{\pi}{8} + \sin^2 \frac{3\pi}{8} + \sin^2 \frac{5\pi}{8} + \sin^2 \frac{7\pi}{8} = 2\]
Prove that: \[\left( \cos \alpha + \cos \beta^2 \right) + \left( \sin \alpha + \sin \beta \right)^2 = 4 \cos^2 \left( \frac{\alpha - \beta}{2} \right)\]
Prove that: \[1 + \cos^2 2x = 2 \left( \cos^4 x + \sin^4 x \right)\]
Prove that:\[\tan\left( \frac{\pi}{4} + x \right) + \tan\left( \frac{\pi}{4} - x \right) = 2 \sec 2x\]
Prove that: \[\cot \frac{\pi}{8} = \sqrt{2} + 1\]
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 \[\sin x = \frac{\sqrt{5}}{3}\] and x lies in IInd quadrant, find the values of \[\cos\frac{x}{2}, \sin\frac{x}{2} \text{ and } \tan \frac{x}{2}\] .
Prove that: \[\cos\frac{\pi}{5}\cos\frac{2\pi}{5}\cos\frac{4\pi}{5}\cos\frac{8\pi}{5} = \frac{- 1}{16}\]
Prove that: \[\cos \frac{\pi}{65} \cos \frac{2\pi}{65} \cos\frac{4\pi}{65} \cos\frac{8\pi}{65} \cos\frac{16\pi}{65} \cos\frac{32\pi}{65} = \frac{1}{64}\]
If \[2 \tan \alpha = 3 \tan \beta,\] prove that \[\tan \left( \alpha - \beta \right) = \frac{\sin 2\beta}{5 - \cos 2\beta}\] .
If \[\sin \alpha + \sin \beta = a \text{ and } \cos \alpha + \cos \beta = b\] , prove that
(ii) \[\cos \left( \alpha - \beta \right) = \frac{a^2 + b^2 - 2}{2}\]
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 \[a \cos2x + b \sin2x = c\] has α and β as its roots, then prove that
(iii)\[\tan\left( \alpha + \beta \right) = \frac{b}{a}\]
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: \[\cos^3 x \sin 3x + \sin^3 x \cos 3x = \frac{3}{4} \sin 4x\]
\[\cot x + \cot\left( \frac{\pi}{3} + x \right) + \cot\left( \frac{2\pi}{3} + x \right) = 3 \cot 3x\]
Prove that: \[\sin^2 \frac{2\pi}{5} - \sin^{2 -} \frac{\pi}{3} = \frac{\sqrt{5} - 1}{8}\]
Prove that: \[\cos\frac{\pi}{15}\cos\frac{2\pi}{15}\cos\frac{4\pi}{15}\cos\frac{7\pi}{15} = \frac{1}{16}\]
Prove that: \[\cos 6° \cos 42° \cos 66° \cos 78° = \frac{1}{16}\]
If \[\frac{\pi}{2} < x < \pi,\] the write the value of \[\sqrt{2 + \sqrt{2 + 2 \cos 2x}}\] in the simplest form.
Write the value of \[\cos^2 76° + \cos^2 16° - \cos 76° \cos 16°\]
If \[\cos x = \frac{1}{2} \left( a + \frac{1}{a} \right),\] and \[\cos 3 x = \lambda \left( a^3 + \frac{1}{a^3} \right)\] then \[\lambda =\]
If \[\tan \alpha = \frac{1 - \cos \beta}{\sin \beta}\] , then
If \[\sin \alpha + \sin \beta = a \text{ and } \cos \alpha - \cos \beta = b \text{ then } \tan \frac{\alpha - \beta}{2} =\]
The value of \[\tan x \sin \left( \frac{\pi}{2} + x \right) \cos \left( \frac{\pi}{2} - x \right)\]
If \[A = 2 \sin^2 x - \cos 2x\] , then A lies in the interval
If \[\tan \left( \pi/4 + x \right) + \tan \left( \pi/4 - x \right) = \lambda \sec 2x, \text{ then } \]
The value of \[\cos^4 x + \sin^4 x - 6 \cos^2 x \sin^2 x\] is
If tanθ = `1/2` and tanΦ = `1/3`, then the value of θ + Φ is ______.
The value of cos12° + cos84° + cos156° + cos132° is ______.
The value of `sin pi/10 sin (13pi)/10` is ______.
`["Hint: Use" sin18^circ = (sqrt5 - 1)/4 "and" cos36^circ = (sqrt5 + 1)/4]`
If A lies in the second quadrant and 3tanA + 4 = 0, then the value of 2cotA – 5cosA + sinA is equal to ______.
If k = `sin(pi/18) sin((5pi)/18) sin((7pi)/18)`, then the numerical value of k is ______.
