Advertisements
Advertisements
प्रश्न
3(sinx – cosx)4 + 6(sinx + cosx)2 + 4(sin6x + cos6x) = ______.
Advertisements
उत्तर
3(sinx – cosx)4 + 6(sinx + cosx)2 + 4(sin6x + cos6x) = 13.
Explanation:
Given expression is 3(sinx – cosx)4 + 6(sinx + cosx)2 + 4(sin6x + cos6x)
= 3[sin2x + cos2x – 2 sinx cosx]2 + 6(sin2x + cos2x + 2sinx cosx) + 4[(sin2x)3 + (cos2x)3]
= 3[1 – 2sinx cosx]2 + 6(1 + 2sinx cosx) + 4[(sin2x + cos2x)3 – 3sin2x cos2x (sin2x + cos2x)]
= 3[1 + 4sin2x cos2x – 4sinx cosx] + 6(1 + 2 sinx cosx) + 4[1 – 3sin2x cos2x]
= 3 + 12sin2x cos2x – 12sinx cosx + 6 + 12sinx cosx + 4 – 12sin2x cos2x
= 3 + 6 + 4
= 13
APPEARS IN
संबंधित प्रश्न
Prove that `cot^2 pi/6 + cosec (5pi)/6 + 3 tan^2 pi/6 = 6`
Prove the following:
sin 2x + 2sin 4x + sin 6x = 4cos2 x sin 4x
Prove the following:
`(cos 4x + cos 3x + cos 2x)/(sin 4x + sin 3x + sin 2x) = cot 3x`
Prove the following:
`tan 4x = (4tan x(1 - tan^2 x))/(1 - 6tan^2 x + tan^4 x)`
Prove that: `((sin 7x + sin 5x) + (sin 9x + sin 3x))/((cos 7x + cos 5x) + (cos 9x + cos 3x)) = tan 6x`
If \[\sin A = \frac{4}{5}\] and \[\cos B = \frac{5}{13}\], where 0 < A, \[B < \frac{\pi}{2}\], find the value of the following:
cos (A − B)
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 \[\sin A = \frac{3}{5}, \cos B = - \frac{12}{13}\], where A and B both lie in second quadrant, find the value of sin (A + B).
If \[\tan A = \frac{3}{4}, \cos B = \frac{9}{41}\], where π < A < \[\frac{3\pi}{2}\] and 0 < B <\[\frac{\pi}{2}\], find tan (A + B).
If \[\sin A = \frac{1}{2}, \cos B = \frac{\sqrt{3}}{2}\], where \[\frac{\pi}{2}\] < A < π and 0 < B < \[\frac{\pi}{2}\], find the following:
tan (A - B)
Evaluate the following:
cos 47° cos 13° − sin 47° sin 13°
Evaluate the following:
sin 36° cos 9° + cos 36° sin 9°
Evaluate the following:
cos 80° cos 20° + sin 80° sin 20°
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)
If \[\tan A = \frac{5}{6}\text{ and }\tan B = \frac{1}{11}\], prove that \[A + B = \frac{\pi}{4}\].
Prove that:
sin2 (n + 1) A − sin2 nA = sin (2n + 1) A sin A.
Prove that:
Prove that:
\[\tan\frac{\pi}{12} + \tan\frac{\pi}{6} + \tan\frac{\pi}{12}\tan\frac{\pi}{6} = 1\]
If x lies in the first quadrant and \[\cos x = \frac{8}{17}\], then prove that:
If α, β are two different values of x lying between 0 and 2π, which satisfy the equation 6 cos x + 8 sin x = 9, find the value of sin (α + β).
If \[\tan\theta = \frac{\sin\alpha - \cos\alpha}{\sin\alpha + \cos\alpha}\] , then show that \[\sin\alpha + \cos\alpha = \sqrt{2}\cos\theta\].
Show that sin 100° − sin 10° is positive.
If a = b \[\cos \frac{2\pi}{3} = c \cos\frac{4\pi}{3}\] then write the value of ab + bc + ca.
If A + B + C = π, then sec A (cos B cos C − sin B sin C) 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 in ∆ABC, tan A + tan B + tan C = 6, then cot A cot B cot C =
If cot (α + β) = 0, sin (α + 2β) is equal to
If tan (A − B) = 1 and sec (A + B) = \[\frac{2}{\sqrt{3}}\], the smallest positive value of B is
If cos (A − B) \[= \frac{3}{5}\] and tan A tan B = 2, then
If α and β are the solutions of the equation a tan θ + b sec θ = c, then show that tan (α + β) = `(2ac)/(a^2 - c^2)`.
Show that 2 sin2β + 4 cos (α + β) sin α sin β + cos 2(α + β) = cos 2α
If `(sin(x + y))/(sin(x - y)) = (a + b)/(a - b)`, then show that `tanx/tany = a/b` [Hint: Use Componendo and Dividendo].
The value of sin(45° + θ) - cos(45° - θ) is ______.
State whether the statement is True or False? Also give justification.
If cosecx = 1 + cotx then x = 2nπ, 2nπ + `pi/2`
State whether the statement is True or False? Also give justification.
If tan(π cosθ) = cot(π sinθ), then `cos(theta - pi/4) = +- 1/(2sqrt(2))`.
