Advertisements
Advertisements
प्रश्न
Prove the following:
cos2 2x – cos2 6x = sin 4x sin 8x
Advertisements
उत्तर
cos2 2x – cos2 6x = (cos2x + cos6x) (cos 2x - cos 6x)
= `2cos ((2x + 6x)/2) cos ((2x - 6 x) /2) - 2sin ((2x - 6x)/2) sin ((2x + 6x)/2)`
= (2cos 4x cos2x) (2sin 4x sin 2x)
= (2sin 2x cos2x)(2sin 4x cos4x) = sin 4x sin 8x
APPEARS IN
संबंधित प्रश्न
Prove the following:
sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n + 2)x = cos x
Prove the following:
`cos ((3pi)/4 + x) - cos((3pi)/4 - x) = -sqrt2 sin x`
Prove the following:
`(cos9x - cos5x)/(sin17x - sin 3x) = - (sin2x)/(cos 10x)`
Prove the following:
`(sin 5x + sin 3x)/(cos 5x + cos 3x) = tan 4x`
Prove the following:
cos 4x = 1 – 8sin2 x cos2 x
If \[\sin A = \frac{12}{13}\text{ and } \sin B = \frac{4}{5}\], where \[\frac{\pi}{2}\] < A < π and 0 < B < \[\frac{\pi}{2}\], find the following:
sin (A + B)
If \[\sin A = \frac{1}{2}, \cos B = \frac{12}{13}\], where \[\frac{\pi}{2}\]< A < π and \[\frac{3\pi}{2}\] < B < 2π, find tan (A − B).
Evaluate the following:
sin 36° cos 9° + cos 36° sin 9°
If \[\cos A = - \frac{12}{13}\text{ and }\cot B = \frac{24}{7}\], where A lies in the second quadrant and B in the third quadrant, find the values of the following:
tan (A + B)
Prove that
Prove that \[\frac{\tan 69^\circ + \tan 66^\circ}{1 - \tan 69^\circ \tan 66^\circ} = - 1\].
Prove that: \[\frac{\sin \left( A + B \right) + \sin \left( A - B \right)}{\cos \left( A + B \right) + \cos \left( A - B \right)} = \tan A\]
Prove that:
\[\frac{\tan^2 2x - \tan^2 x}{1 - \tan^2 2x \tan^2 x} = \tan 3x \tan x\]
Prove that sin2 (n + 1) A − sin2 nA = sin (2n + 1) A sin A.
If x lies in the first quadrant and \[\cos x = \frac{8}{17}\], then prove that:
If tan α = x +1, tan β = x − 1, show that 2 cot (α − β) = x2.
Prove that \[\left( 2\sqrt{3} + 3 \right) \sin x + 2\sqrt{3} \cos x\] lies between \[- \left( 2\sqrt{3} + \sqrt{15} \right) \text{ and } \left( 2\sqrt{3} + \sqrt{15} \right)\]
If 12 sin x − 9sin2 x attains its maximum value at x = α, then write the value of sin α.
Write the interval in which the value of 5 cos x + 3 cos \[\left( x + \frac{\pi}{3} \right) + 3\] lies.
If \[\frac{\cos \left( x - y \right)}{\cos \left( x + y \right)} = \frac{m}{n}\] then write the value of tan x tan y.
tan 20° + tan 40° + \[\sqrt{3}\] tan 20° tan 40° is equal to
If \[\tan A = \frac{a}{a + 1}\text{ and } \tan B = \frac{1}{2a + 1}\]
If 3 sin x + 4 cos x = 5, then 4 sin x − 3 cos x =
If cot (α + β) = 0, sin (α + 2β) is equal to
If tan θ1 tan θ2 = k, then \[\frac{\cos \left( \theta_1 - \theta_2 \right)}{\cos \left( \theta_1 + \theta_2 \right)} =\]
If \[\tan\theta = \frac{1}{2}\] and \[\tan\phi = \frac{1}{3}\], then the value of \[\tan\phi = \frac{1}{3}\] is
If tan (π/4 + x) + tan (π/4 − x) = a, then tan2 (π/4 + x) + tan2 (π/4 − x) =
If tan 69° + tan 66° − tan 69° tan 66° = 2k, then k =
If 3 tan (θ – 15°) = tan (θ + 15°), 0° < θ < 90°, then θ = ______.
If sinθ + cosθ = 1, then find the general value of θ.
If sin(θ + α) = a and sin(θ + β) = b, then prove that cos 2(α - β) - 4ab cos(α - β) = 1 - 2a2 - 2b2
[Hint: Express cos(α - β) = cos((θ + α) - (θ + β))]
If cos(θ + Φ) = m cos(θ – Φ), then prove that 1 tan θ = `(1 - m)/(1 + m) cot phi`
[Hint: Express `(cos(theta + Φ))/(cos(theta - Φ)) = m/1` and apply Componendo and Dividendo]
Find the general solution of the equation `(sqrt(3) - 1) costheta + (sqrt(3) + 1) sin theta` = 2
[Hint: Put `sqrt(3) - 1` = r sinα, `sqrt(3) + 1` = r cosα which gives tanα = `tan(pi/4 - pi/6)` α = `pi/12`]
The value of tan3A - tan2A - tanA is equal to ______.
If α + β = `pi/4`, then the value of (1 + tan α)(1 + tan β) is ______.
3(sinx – cosx)4 + 6(sinx + cosx)2 + 4(sin6x + cos6x) = ______.
In the following match each item given under the column C1 to its correct answer given under the column C2:
| Column A | Column B |
| (a) sin(x + y) sin(x – y) | (i) cos2x – sin2y |
| (b) cos (x + y) cos (x – y) | (ii) `(1 - tan theta)/(1 + tan theta)` |
| (c) `cot(pi/4 + theta)` | (iii) `(1 + tan theta)/(1 - tan theta)` |
| (d) `tan(pi/4 + theta)` | (iv) sin2x – sin2y |
