Advertisements
Advertisements
प्रश्न
Prove that: \[\cos 4x = 1 - 8 \cos^2 x + 8 \cos^4 x\]
Advertisements
उत्तर १
\[LHS = \cos4x\]
\[ = \cos\left( 2 \times 2x \right)\]
\[ = 2 \cos^2 \times 2x - 1 \left[ \because \cos2\theta = 2 \cos^2 \theta - 1 \right]\]
\[ = 2 \left( 2 \cos^2 x - 1 \right)^2 - 1 \left[ \because \cos^2 2\theta = \left( 2 \cos^2 \theta - 1 \right)^2 \right]\]
\[= 2\left( 4 \cos^4 x - 4 \cos^2 x + 1 \right) - 1\]
\[ = 8 \cos^4 x - 8 \cos^2 x + 1\]
\[ = 1 - 8 \cos^2 x + 8 \cos^4 x = RHS\]
\[Hence proved .\]
उत्तर २
\[LHS = \cos4x\]
\[ = \cos\left( 2 \times 2x \right)\]
\[ = 2 \cos^2 \times 2x - 1 \left[ \because \cos2\theta = 2 \cos^2 \theta - 1 \right]\]
\[ = 2 \left( 2 \cos^2 x - 1 \right)^2 - 1 \left[ \because \cos^2 2\theta = \left( 2 \cos^2 \theta - 1 \right)^2 \right]\]
\[= 2\left( 4 \cos^4 x - 4 \cos^2 x + 1 \right) - 1\]
\[ = 8 \cos^4 x - 8 \cos^2 x + 1\]
\[ = 1 - 8 \cos^2 x + 8 \cos^4 x = RHS\]
\[\text{ Hence proved } .\]
APPEARS IN
संबंधित प्रश्न
Prove that: \[\sqrt{\frac{1 - \cos 2x}{1 + \cos 2x}} = \tan x\]
Prove that: \[\frac{\cos x}{1 - \sin x} = \tan \left( \frac{\pi}{4} + \frac{x}{2} \right)\]
Prove that: \[\sin 4x = 4 \sin x \cos^3 x - 4 \cos x \sin^3 x\]
Prove that \[\sin 3x + \sin 2x - \sin x = 4 \sin x \cos\frac{x}{2} \cos\frac{3x}{2}\]
Prove that: \[\cos 7° \cos 14° \cos 28° \cos 56°= \frac{\sin 68°}{16 \cos 83°}\]
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 \[\sec \left( x + \alpha \right) + \sec \left( x - \alpha \right) = 2 \sec x\] , prove that \[\cos x = \pm \sqrt{2} \cos\frac{\alpha}{2}\]
If \[\cos \alpha + \cos \beta = \frac{1}{3}\] and sin \[\sin\alpha + \sin \beta = \frac{1}{4}\] , prove that \[\cos\frac{\alpha - \beta}{2} = \pm \frac{5}{24}\]
\[\cot x + \cot\left( \frac{\pi}{3} + x \right) + \cot\left( \frac{2\pi}{3} + x \right) = 3 \cot 3x\]
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 \[\tan\frac{x}{2} = \frac{m}{n}\] , then write the value of m sin x + n cos x.
If \[\frac{\pi}{2} < x < \frac{3\pi}{2}\] , then write the value of \[\sqrt{\frac{1 + \cos 2x}{2}}\]
If \[\pi < x < \frac{3\pi}{2}\], then write the value of \[\sqrt{\frac{1 - \cos 2x}{1 + \cos 2x}}\] .
Write the value of \[\cos^2 76° + \cos^2 16° - \cos 76° \cos 16°\]
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 in a \[∆ ABC, \tan A + \tan B + \tan C = 0\], then
If \[2 \tan \alpha = 3 \tan \beta, \text{ then } \tan \left( \alpha - \beta \right) =\]
If \[\tan \alpha = \frac{1 - \cos \beta}{\sin \beta}\] , then
If \[5 \sin \alpha = 3 \sin \left( \alpha + 2 \beta \right) \neq 0\] , then \[\tan \left( \alpha + \beta \right)\] is equal to
If \[A = 2 \sin^2 x - \cos 2x\] , then A lies in the interval
The value of \[\frac{\cos 3x}{2 \cos 2x - 1}\] is equal to
The value of \[\tan x \tan \left( \frac{\pi}{3} - x \right) \tan \left( \frac{\pi}{3} + x \right)\] is
The value of `cos^2 48^@ - sin^2 12^@` is ______.
The value of `cos pi/5 cos (2pi)/5 cos (4pi)/5 cos (8pi)/5` is ______.
If tanθ + sinθ = m and tanθ – sinθ = n, then prove that m2 – n2 = 4sinθ tanθ
[Hint: m + n = 2tanθ, m – n = 2sinθ, then use m2 – n2 = (m + n)(m – n)]
If sinθ = `(-4)/5` and θ lies in the third quadrant then the value of `cos theta/2` is ______.
The value of `sin pi/18 + sin pi/9 + sin (2pi)/9 + sin (5pi)/18` is given by ______.
The value of cos248° – sin212° is ______.
[Hint: Use cos2A – sin2 B = cos(A + B) cos(A – B)]
The value of `(sin 50^circ)/(sin 130^circ)` is ______.
If k = `sin(pi/18) sin((5pi)/18) sin((7pi)/18)`, then the numerical value of k is ______.
