Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[LHS = \sin^3 x + \sin^3 \left( \frac{2\pi}{3} + x \right) + \sin^3 \left( \frac{4\pi}{3} + x \right)\]
\[ = \frac{3\text{ sin } x - \sin3x}{4} + \frac{3\sin\left( \frac{2\pi}{3} + x \right) - \sin3\left( \frac{2\pi}{3} + x \right)}{4} + \frac{3\sin\left( \frac{4\pi}{3} + x \right) - \sin3\left( \frac{4\pi}{3} + x \right)}{4} \]
\[ \left[ \sin^3 \theta = \frac{3sin\theta - \sin3\theta}{4} \right]\]
\[ = \frac{3\text{ sin } x - \sin3x}{4} + \frac{3\sin\left\{ \pi - \left( \frac{2\pi}{3} + x \right) \right\} - \sin\left( 2\pi + 3x \right)}{4} + \frac{3\sin\left\{ \pi + \left( \frac{\pi}{3} + x \right) \right\} - \sin\left( 4\pi + 3x \right)}{4}\]
\[ = \frac{1}{4}\left[ \left( 3\text{ sin } x - \sin3x \right) + \left\{ 3\sin\left( \frac{\pi}{3} - x \right) - \sin3x \right\} - \left\{ 3\sin\left( \frac{\pi}{3} + x \right) + \sin3x \right\} \right]\]
\[ = \frac{1}{4}\left[ 3\text{ sin } x - \sin3x + 3\sin\left( \frac{\pi}{3} - x \right) - 3\sin\left( \frac{\pi}{3} + x \right) - \sin3x - \sin3x \right]\]
\[= \frac{1}{4}\left[ 3\text{ sin } x - 3\sin3x + 3\left\{ \sin\left( \frac{\pi}{3} - x \right) - \sin\left( \frac{\pi}{3} + x \right) \right\} \right]\]
\[ = \frac{1}{4}\left[ 3\text{ sin } x - 3\sin3x + 3\left\{ 2\cos\frac{\frac{\pi}{3} - x + \frac{\pi}{3} + x}{2}\sin\frac{\frac{\pi}{3} - x - \frac{\pi}{3} - x}{2} \right\} \right]\]
\[ \left[ \because sinC - sinD = 2\cos\frac{C + D}{2}\sin\frac{C - D}{2} \right]\]
\[ = \frac{1}{4}\left[ 3\text{ sin } x - 3\sin3x + 6\cos\frac{\pi}{3}\sin\left( - x \right) \right]\]
\[ = \frac{1}{4}\left[ 3\text{ sin } x - 3\text{ sin } 3x - 3\text{ sin } x \right]\]
\[ = - \frac{3}{4}\text{ sin } x\]
\[ = RHS\]
\[\text{ Hence proved } .\]
APPEARS IN
संबंधित प्रश्न
Prove that: \[\frac{\sin 2x}{1 - \cos 2x} = cot x\]
Prove that: \[\frac{\sin 2x}{1 + \cos 2x} = \tan x\]
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: \[\left( \cos \alpha + \cos \beta^2 \right) + \left( \sin \alpha + \sin \beta \right)^2 = 4 \cos^2 \left( \frac{\alpha - \beta}{2} \right)\]
Prove that: \[\sin^2 \left( \frac{\pi}{8} + \frac{x}{2} \right) - \sin^2 \left( \frac{\pi}{8} - \frac{x}{2} \right) = \frac{1}{\sqrt{2}} \sin x\]
Prove that: \[\left( \sin 3x + \sin x \right) \sin x + \left( \cos 3x - \cos x \right) \cos x = 0\]
Prove that: \[\sin 4x = 4 \sin x \cos^3 x - 4 \cos x \sin^3 x\]
If 0 ≤ x ≤ π and x lies in the IInd quadrant such that \[\sin x = \frac{1}{4}\]. Find the values of \[\cos\frac{x}{2}, \sin\frac{x}{2} \text{ and } \tan\frac{x}{2}\]
If \[\cos x = \frac{4}{5}\] and x is acute, find tan 2x
Prove that: \[\cos 7° \cos 14° \cos 28° \cos 56°= \frac{\sin 68°}{16 \cos 83°}\]
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 \[2 \tan\frac{\alpha}{2} = \tan\frac{\beta}{2}\] , prove that \[\cos \alpha = \frac{3 + 5 \cos \beta}{5 + 3 \cos \beta}\]
If \[\cos \alpha + \cos \beta = \frac{1}{3}\] and sin \[\sin\alpha + \sin \beta = \frac{1}{4}\] , prove that \[\cos\frac{\alpha - \beta}{2} = \pm \frac{5}{24}\]
Prove that \[\left| \sin x \sin \left( \frac{\pi}{3} - x \right) \sin \left( \frac{\pi}{3} + x \right) \right| \leq \frac{1}{4}\] for all values of x
Prove that \[\left| \cos x \cos \left( \frac{\pi}{3} - x \right) \cos \left( \frac{\pi}{3} + x \right) \right| \leq \frac{1}{4}\] for all values of x
Prove that: \[\sin^2 24°- \sin^2 6° = \frac{\sqrt{5} - 1}{8}\]
Write the value of \[\cos^2 76° + \cos^2 16° - \cos 76° \cos 16°\]
If \[\text{ sin } x + \text{ cos } x = a\], find the value of \[\left|\text { sin } x - \text{ cos } x \right|\] .
The value of \[\cos \frac{\pi}{65} \cos \frac{2\pi}{65} \cos \frac{4\pi}{65} \cos \frac{8\pi}{65} \cos \frac{16\pi}{65} \cos \frac{32\pi}{65}\] is
If \[\cos 2x + 2 \cos x = 1\] then, \[\left( 2 - \cos^2 x \right) \sin^2 x\] is equal to
For all real values of x, \[\cot x - 2 \cot 2x\] is equal to
If in a \[∆ ABC, \tan A + \tan B + \tan C = 0\], then
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 \[A = 2 \sin^2 x - \cos 2x\] , then A lies in the interval
The value of \[\cos^4 x + \sin^4 x - 6 \cos^2 x \sin^2 x\] is
The value of \[\tan x \tan \left( \frac{\pi}{3} - x \right) \tan \left( \frac{\pi}{3} + x \right)\] is
The value of \[\tan x + \tan \left( \frac{\pi}{3} + x \right) + \tan \left( \frac{2\pi}{3} + x \right)\] is
The value of \[\frac{\sin 5 \alpha - \sin 3\alpha}{\cos 5 \alpha + 2 \cos 4\alpha + \cos 3\alpha} =\]
The value of sin 20° sin 40° sin 60° sin 80° is ______.
Prove that sin 4A = 4sinA cos3A – 4 cosA sin3A
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).
Find the value of the expression `cos^4 pi/8 + cos^4 (3pi)/8 + cos^4 (5pi)/8 + cos^4 (7pi)/8`
[Hint: Simplify the expression to `2(cos^4 pi/8 + cos^4 (3pi)/8) = 2[(cos^2 pi/8 + cos^2 (3pi)/8)^2 - 2cos^2 pi/8 cos^2 (3pi)/8]`
If sinθ = `(-4)/5` and θ lies in the third quadrant then the value of `cos theta/2` is ______.
If tanA = `(1 - cos "B")/sin"B"`, then tan2A = ______.
