Advertisements
Advertisements
प्रश्न
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))]
Advertisements
उत्तर
sin(θ + α) = a and sin(θ + β) = b
L.H.S = cos 2(α - β) - 4ab cos(α - β)
Using cos2x = 2cos2x - 1,
Let us solve,
⇒ LHS = 2cos2(α - β) - 1 - 4ab cos(α - β)
⇒ LHS = 2cos(α - β) {cos(α - β) - 2ab} - 1
Since,
cos(α - β) = cos{(θ + α) - (θ + β)}
cos(A - B) = cosA cosB + sinA sinB
⇒ cos(α - β) = cos(θ + α) cos(θ + β) + sin(θ + α) sin(θ + β)
Since, sin(θ + α) = a
⇒ cos(θ + α) = `sqrt(1 – sin^2(θ + alpha))`
= `sqrt(1 – "a"^2)`
Similarly,
cos(θ + β) = `sqrt(1 – b^2)`
Therefore,
cos(α - β) = `sqrt(1 - a^2) sqrt(1 - b^2) + ab`
Therefore,
L.H.S = `2{ab + sqrt(1 – a^2)(1 – b^2)}{ab + sqrt(1 – a^2)(1 – b^2) - 2ab} – 1`
⇒ L.H.S =`2{sqrt(1 – a^2)(1 – b^2) + ab}{sqrt(1 – a^2)(1 – b^2) – ab} - 1`
Using (x + y)(x - y) = x2 - y2
⇒ L.H.S = 2{(1 - a2)(1 - b2) - a2b2} - 1
⇒ L.H.S = 2{1 - a2 - b2 + a2b2} - 1
⇒ L.H.S = 2 - 2a2 - 2b2 - 1
⇒ L.H.S = 1 - 2a2 - 2b2 = RHS
Therefore,
We get,
cos 2(α - β) - 4ab cos(α - β) = 1 - 2a2 - 2b2.
APPEARS IN
संबंधित प्रश्न
Prove that `cot^2 pi/6 + cosec (5pi)/6 + 3 tan^2 pi/6 = 6`
Prove the following: `(tan(pi/4 + x))/(tan(pi/4 - x)) = ((1+ tan x)/(1- tan x))^2`
Prove the following:
`(sin x - sin 3x)/(sin^2 x - cos^2 x) = 2sin x`
Prove the following:
cos 6x = 32 cos6 x – 48 cos4 x + 18 cos2 x – 1
If \[\cos A = - \frac{24}{25}\text{ and }\cos B = \frac{3}{5}\], where π < A < \[\frac{3\pi}{2}\text{ and }\frac{3\pi}{2}\]< B < 2π, find the following:
sin (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 80° cos 20° + sin 80° sin 20°
Prove that:
\[\cos^2 45^\circ - \sin^2 15^\circ = \frac{\sqrt{3}}{4}\]
Prove that:
sin2 (n + 1) A − sin2 nA = sin (2n + 1) A sin A.
Prove that:
Prove that:
cos2 A + cos2 B − 2 cos A cos B cos (A + B) = sin2 (A + B)
Prove that:
tan 13x − tan 9x − tan 4x = tan 13x tan 9x tan 4x
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).
Prove that:
\[\frac{1}{\sin \left( x - a \right) \sin \left( x - b \right)} = \frac{\cot \left( x - a \right) - \cot \left( x - b \right)}{\sin \left( a - b \right)}\]
If tan α = x +1, tan β = x − 1, show that 2 cot (α − β) = x2.
If angle \[\theta\] is divided into two parts such that the tangents of one part is \[\lambda\] times the tangent of other, and \[\phi\] is their difference, then show that\[\sin\theta = \frac{\lambda + 1}{\lambda - 1}\sin\phi\]
Find the maximum and minimum values of each of the following trigonometrical expression:
12 sin x − 5 cos x
Write the maximum value of 12 sin x − 9 sin2 x.
Write the interval in which the value of 5 cos x + 3 cos \[\left( x + \frac{\pi}{3} \right) + 3\] lies.
If A + B + C = π, then sec A (cos B cos C − sin B sin C) is equal to
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 \[\cos P = \frac{1}{7}\text{ and }\cos Q = \frac{13}{14}\], where P and Q both are acute angles. Then, the value of P − Q is
If \[\tan\alpha = \frac{x}{x + 1}\] and \[\tan\alpha = \frac{x}{x + 1}\], then \[\alpha + \beta\] is equal to
Show that 2 sin2β + 4 cos (α + β) sin α sin β + cos 2(α + β) = cos 2α
Find the most general value of θ satisfying the equation tan θ = –1 and cos θ = `1/sqrt(2)`.
If f(x) = cos2x + sec2x, then ______.
[Hint: A.M ≥ G.M.]
If α + β = `pi/4`, then the value of (1 + tan α)(1 + tan β) is ______.
If tanθ = `a/b`, then bcos2θ + asin2θ is equal to ______.
3(sinx – cosx)4 + 6(sinx + cosx)2 + 4(sin6x + cos6x) = ______.
Given x > 0, the values of f(x) = `-3cos sqrt(3 + x + x^2)` lie in the interval ______.
