Advertisements
Advertisements
प्रश्न
Prove that: \[\sin^2 24°- \sin^2 6° = \frac{\sqrt{5} - 1}{8}\]
Advertisements
उत्तर
\[LHS = \sin^2 24° - \sin^2 6° \]
\[ = \sin\left( 24° + 6° \right) \sin\left( 24° - 6° \right) \left[ \sin\left( A + B \right) \sin\left( A - B \right) = \sin^2 A - \sin^2 B \right]\]
\[ = \sin30° \sin18° \]
\[ = \frac{1}{2} \times \frac{\sqrt{5} - 1}{4} \left( \because \sin18° = \frac{\sqrt{5} - 1}{4} \right)\]
\[ = \frac{\sqrt{5} - 1}{8}\]
\[ = RHS\]
\[\text{ Hence proved } .\]
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: \[\left( \cos \alpha + \cos \beta^2 \right) + \left( \sin \alpha + \sin \beta \right)^2 = 4 \cos^2 \left( \frac{\alpha - \beta}{2} \right)\]
Show that: \[2 \left( \sin^6 x + \cos^6 x \right) - 3 \left( \sin^4 x + \cos^4 x \right) + 1 = 0\]
If \[\cos x = - \frac{3}{5}\] and x lies in IInd quadrant, find the values of sin 2x and \[\sin\frac{x}{2}\] .
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 \[\tan A = \frac{1}{7}\] 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 \[\sin \alpha + \sin \beta = a \text{ and } \cos \alpha + \cos \beta = b\] , 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 \[\cos\frac{\alpha - \beta}{2} = \pm \frac{5}{24}\]
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: \[\sin^2 \frac{2\pi}{5} - \sin^{2 -} \frac{\pi}{3} = \frac{\sqrt{5} - 1}{8}\]
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 \[\frac{\pi}{2} < x < \pi,\] the write the value of \[\sqrt{2 + \sqrt{2 + 2 \cos 2x}}\] in the simplest form.
In a right angled triangle ABC, write the value of sin2 A + Sin2 B + Sin2 C.
If \[\text{ tan } A = \frac{1 - \text{ cos } B}{\text{ sin } B}\]
, then find the value of tan2A.
If \[\text{ sin } x + \text{ cos } x = a\], find the value of \[\left|\text { sin } x - \text{ cos } x \right|\] .
The value of \[2 \tan \frac{\pi}{10} + 3 \sec \frac{\pi}{10} - 4 \cos \frac{\pi}{10}\] is
If in a \[∆ ABC, \tan A + \tan B + \tan C = 0\], then
If \[\sin \alpha + \sin \beta = a \text{ and } \cos \alpha - \cos \beta = b \text{ then } \tan \frac{\alpha - \beta}{2} =\]
The value of \[\left( \cot \frac{x}{2} - \tan \frac{x}{2} \right)^2 \left( 1 - 2 \tan x \cot 2 x \right)\] is
\[\sin^2 \left( \frac{\pi}{18} \right) + \sin^2 \left( \frac{\pi}{9} \right) + \sin^2 \left( \frac{7\pi}{18} \right) + \sin^2 \left( \frac{4\pi}{9} \right) =\]
If \[5 \sin \alpha = 3 \sin \left( \alpha + 2 \beta \right) \neq 0\] , then \[\tan \left( \alpha + \beta \right)\] is equal to
\[2 \text{ cos } x - \ cos 3x - \cos 5x - 16 \cos^3 x \sin^2 x\]
The value of \[\frac{\cos 3x}{2 \cos 2x - 1}\] is equal to
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 α and β are acute angles satisfying \[\cos 2 \alpha = \frac{3 \cos 2 \beta - 1}{3 - \cos 2 \beta}\] , then tan α =
Prove that sin 4A = 4sinA cos3A – 4 cosA sin3A
If tan(A + B) = p, tan(A – B) = q, then show that tan 2A = `(p + q)/(1 - pq)`
The value of `(1 - tan^2 15^circ)/(1 + tan^2 15^circ)` is ______.
