Advertisements
Advertisements
प्रश्न
The value of \[\cos^2 \left( \frac{\pi}{6} + x \right) - \sin^2 \left( \frac{\pi}{6} - x \right)\] is
पर्याय
\[\frac{1}{2} \cos 2x\]
0
\[- \frac{1}{2} \cos 2x\]
\[\frac{1}{2}\]
Advertisements
उत्तर
\[\frac{1}{2} \cos 2x\]
\[\text{ We have, } \]
\[ \cos^2 \left( \frac{\pi}{6} + x \right) - \sin^2 \left( \frac{\pi}{6} - x \right)\]
\[ = \cos^2 \left( \frac{\pi}{6} + x \right) - \cos^2 \left[ \frac{\pi}{2} - \left( \frac{\pi}{6} - x \right) \right]\]
\[ = \cos^2 \left( \frac{\pi}{6} + x \right) - \cos^2 \left( \frac{\pi}{3} + x \right)\]
\[ = \left[ \cos\left( \frac{\pi}{6} + x \right) + \cos\left( \frac{\pi}{3} + x \right) \right]\left[ \cos\left( \frac{\pi}{6} + x \right) - \cos\left( \frac{\pi}{3} + x \right) \right]\]
\[ = 2\cos\left( \frac{\frac{\pi}{6} + x + \frac{\pi}{3} + x}{2} \right) \cos\left( \frac{\frac{\pi}{6} + x - \frac{\pi}{3} - x}{2} \right) 2\sin\left( \frac{\frac{\pi}{6} + x + \frac{\pi}{3} + x}{2} \right) \sin\left( \frac{\frac{\pi}{3} + x - \frac{\pi}{6} - x}{2} \right)\]
\[ = 4\cos\left( \frac{\pi}{4} + x \right)\cos\left( - \frac{\pi}{12} \right) \sin\left( \frac{\pi}{4} + x \right) \sin\left( \frac{\pi}{12} \right)\]
\[ = 4\cos\left( \frac{\pi}{4} + x \right)\cos\left( \frac{\pi}{12} \right) \sin\left( \frac{\pi}{4} + x \right) \sin\left( \frac{\pi}{12} \right)\]
\[ = \left[ 2\sin\left( \frac{\pi}{4} + x \right)\cos\left( \frac{\pi}{4} + x \right) \right]\left[ 2 \sin\left( \frac{\pi}{12} \right)\cos\left( \frac{\pi}{12} \right) \right]\]
\[ = \sin\left( \frac{\pi}{2} + 2x \right)\sin\frac{\pi}{6}\]
\[ = \cos2x \times \frac{1}{2}\]
\[ = \frac{1}{2}\cos2x\]
APPEARS IN
संबंधित प्रश्न
Prove that: \[\sqrt{\frac{1 - \cos 2x}{1 + \cos 2x}} = \tan x\]
Prove that: \[\frac{\sin 2x}{1 - \cos 2x} = cot x\]
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: \[\cos^2 \left( \frac{\pi}{4} - x \right) - \sin^2 \left( \frac{\pi}{4} - x \right) = \sin 2x\]
Prove that: \[\sin 4x = 4 \sin x \cos^3 x - 4 \cos x \sin^3 x\]
Prove that: \[\cos^6 A - \sin^6 A = \cos 2A\left( 1 - \frac{1}{4} \sin^2 2A \right)\]
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 \[\tan A = \frac{1}{7}\] and \[\tan B = \frac{1}{3}\] , show that cos 2A = sin 4B
Prove that: \[\cos\frac{\pi}{5}\cos\frac{2\pi}{5}\cos\frac{4\pi}{5}\cos\frac{8\pi}{5} = \frac{- 1}{16}\]
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
(iii)\[\tan\left( \alpha + \beta \right) = \frac{b}{a}\]
Prove that: \[\cos 36° \cos 42° \cos 60° \cos 78° = \frac{1}{16}\]
Prove that: \[\sin\frac{\pi}{5}\sin\frac{2\pi}{5}\sin\frac{3\pi}{5}\sin\frac{4\pi}{5} = \frac{5}{16}\]
Write the value of \[\cos^2 76° + \cos^2 16° - \cos 76° \cos 16°\]
If \[\text{ tan } A = \frac{1 - \text{ cos } B}{\text{ sin } B}\]
, then find the value of tan2A.
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 \[\tan \alpha = \frac{1 - \cos \beta}{\sin \beta}\] , 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 α and β are acute angles satisfying \[\cos 2 \alpha = \frac{3 \cos 2 \beta - 1}{3 - \cos 2 \beta}\] , then tan α =
If \[\tan \frac{x}{2} = \frac{\sqrt{1 - e}}{1 + e} \tan \frac{\alpha}{2}\] , then \[\cos \alpha =\]
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 \[\frac{\sin 5 \alpha - \sin 3\alpha}{\cos 5 \alpha + 2 \cos 4\alpha + \cos 3\alpha} =\]
If \[n = 1, 2, 3, . . . , \text{ then } \cos \alpha \cos 2 \alpha \cos 4 \alpha . . . \cos 2^{n - 1} \alpha\] is equal to
If \[\tan\alpha = \frac{1}{7}, \tan\beta = \frac{1}{3}\], then
\[\cos2\alpha\] is equal to
If A = cos2θ + sin4θ for all values of θ, then prove that `3/4` ≤ A ≤ 1.
The value of `cos pi/5 cos (2pi)/5 cos (4pi)/5 cos (8pi)/5` is ______.
If tan(A + B) = p, tan(A – B) = q, then show that tan 2A = `(p + q)/(1 - pq)`
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)]`.
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 cos12° + cos84° + cos156° + cos132° is ______.
The value of `sin pi/18 + sin pi/9 + sin (2pi)/9 + sin (5pi)/18` is given by ______.
If tanA = `(1 - cos "B")/sin"B"`, then tan2A = ______.
