Advertisements
Advertisements
Question
Prove that: \[\cos 4x = 1 - 8 \cos^2 x + 8 \cos^4 x\]
Advertisements
Solution 1
\[LHS = \cos4x\]
\[ = \cos\left( 2 \times 2x \right)\]
\[ = 2 \cos^2 \times 2x - 1 \left[ \because \cos2\theta = 2 \cos^2 \theta - 1 \right]\]
\[ = 2 \left( 2 \cos^2 x - 1 \right)^2 - 1 \left[ \because \cos^2 2\theta = \left( 2 \cos^2 \theta - 1 \right)^2 \right]\]
\[= 2\left( 4 \cos^4 x - 4 \cos^2 x + 1 \right) - 1\]
\[ = 8 \cos^4 x - 8 \cos^2 x + 1\]
\[ = 1 - 8 \cos^2 x + 8 \cos^4 x = RHS\]
\[Hence proved .\]
Solution 2
\[LHS = \cos4x\]
\[ = \cos\left( 2 \times 2x \right)\]
\[ = 2 \cos^2 \times 2x - 1 \left[ \because \cos2\theta = 2 \cos^2 \theta - 1 \right]\]
\[ = 2 \left( 2 \cos^2 x - 1 \right)^2 - 1 \left[ \because \cos^2 2\theta = \left( 2 \cos^2 \theta - 1 \right)^2 \right]\]
\[= 2\left( 4 \cos^4 x - 4 \cos^2 x + 1 \right) - 1\]
\[ = 8 \cos^4 x - 8 \cos^2 x + 1\]
\[ = 1 - 8 \cos^2 x + 8 \cos^4 x = RHS\]
\[\text{ Hence proved } .\]
APPEARS IN
RELATED QUESTIONS
Prove that \[\sin 3x + \sin 2x - \sin x = 4 \sin x \cos\frac{x}{2} \cos\frac{3x}{2}\]
If \[\cos x = \frac{4}{5}\] and x is acute, find tan 2x
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 \[2 \tan\frac{\alpha}{2} = \tan\frac{\beta}{2}\] , prove that \[\cos \alpha = \frac{3 + 5 \cos \beta}{5 + 3 \cos \beta}\]
If \[a \cos2x + b \sin2x = c\] has α and β as its roots, then prove that
(ii) \[\tan\alpha \tan\beta = \frac{c - a}{c + a}\]
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: \[\sin^2 24°- \sin^2 6° = \frac{\sqrt{5} - 1}{8}\]
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: \[\sin\frac{\pi}{5}\sin\frac{2\pi}{5}\sin\frac{3\pi}{5}\sin\frac{4\pi}{5} = \frac{5}{16}\]
Prove that: \[\cos\frac{\pi}{15} \cos \frac{2\pi}{15} \cos \frac{3\pi}{15} \cos \frac{4\pi}{15} \cos \frac{5\pi}{15} \cos\frac{6\pi}{15} \cos \frac{7\pi}{15} = \frac{1}{128}\]
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 \[\frac{\pi}{2} < x < \pi\], then write the value of \[\frac{\sqrt{1 - \cos 2x}}{1 + \cos 2x}\] .
If \[\pi < x < \frac{3\pi}{2}\], then write the value of \[\sqrt{\frac{1 - \cos 2x}{1 + \cos 2x}}\] .
In a right angled triangle ABC, write the value of sin2 A + Sin2 B + Sin2 C.
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 \[\tan \alpha = \frac{1 - \cos \beta}{\sin \beta}\] , then
If \[\sin \alpha + \sin \beta = a \text{ and } \cos \alpha - \cos \beta = b \text{ then } \tan \frac{\alpha - \beta}{2} =\]
\[2 \text{ cos } x - \ cos 3x - \cos 5x - 16 \cos^3 x \sin^2 x\]
The value of \[2 \sin^2 B + 4 \cos \left( A + B \right) \sin A \sin B + \cos 2 \left( A + B \right)\] is
The value of \[\frac{2\left( \sin 2x + 2 \cos^2 x - 1 \right)}{\cos x - \sin x - \cos 3x + \sin 3x}\] is
\[2 \left( 1 - 2 \sin^2 7x \right) \sin 3x\] is equal to
If \[\tan x = t\] then \[\tan 2x + \sec 2x =\]
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 ______.
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)]
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]`
The value of `(1 - tan^2 15^circ)/(1 + tan^2 15^circ)` is ______.
The value of cos12° + cos84° + cos156° + cos132° is ______.
The value of cos248° – sin212° is ______.
[Hint: Use cos2A – sin2 B = cos(A + B) cos(A – B)]
If tanA = `(1 - cos "B")/sin"B"`, then tan2A = ______.
