Advertisements
Advertisements
Question
Prove that: `(cos x - cosy)^2 + (sin x - sin y)^2 = 4 sin^2 (x - y)/2`
Advertisements
Solution
L.H.S. = (cos x – cos y)2 + (sin x – sin y)2
= `( -2sin (x + y)/2 sin (x - y)/2)^2 + (2cos (x + y)/2 sin (x - y)/2)^2`
= `4sin^2 (x +y)/2 sin^2 (x - y)/2 + 4cos^2 (x +y)/2 sin^2 (x - y)/2`
= `4sin^2 (x -y) [ sin^2 (x + y)/2 + cos^2 (x +y)/2]`
= `4sin^2 (x - y)/2` `[∵ sin^2 (x + y)/2 + cos^2 (x +y)/2 = 1]`
= R.H.S.
APPEARS IN
RELATED QUESTIONS
Prove that `2 sin^2 pi/6 + cosec^2 (7pi)/6 cos^2 pi/3 = 3/2`
Prove the following:
sin2 6x – sin2 4x = sin 2x sin 10x
Prove the following:
cos2 2x – cos2 6x = sin 4x sin 8x
Prove the following:
`(cos9x - cos5x)/(sin17x - sin 3x) = - (sin2x)/(cos 10x)`
Prove the following:
`(sin x - siny)/(cos x + cos y)= tan (x -y)/2`
Prove the following:
cos 6x = 32 cos6 x – 48 cos4 x + 18 cos2 x – 1
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:
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:
sin (A − B)
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:
cos (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:
sin 78° cos 18° − cos 78° sin 18°
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:
sin (A + B)
Prove that:
\[\frac{7\pi}{12} + \cos\frac{\pi}{12} = \sin\frac{5\pi}{12} - \sin\frac{\pi}{12}\]
Prove that
If \[\tan A = \frac{5}{6}\text{ and }\tan B = \frac{1}{11}\], prove that \[A + B = \frac{\pi}{4}\].
Prove that:
sin2 B = sin2 A + sin2 (A − B) − 2 sin A cos B sin (A − B)
Prove that:
tan 8x − tan 6x − tan 2x = tan 8x tan 6x tan 2x
Prove that:
tan 36° + tan 9° + tan 36° tan 9° = 1
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 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 sin α + sin β = a and cos α + cos β = b, show that
Prove that:
Write the maximum value of 12 sin x − 9 sin2 x.
If a = b \[\cos \frac{2\pi}{3} = c \cos\frac{4\pi}{3}\] then write the value of ab + bc + ca.
The value of \[\cos^2 \left( \frac{\pi}{6} + x \right) - \sin^2 \left( \frac{\pi}{6} - x \right)\] is
If \[\tan\alpha = \frac{x}{x + 1}\] and \[\tan\alpha = \frac{x}{x + 1}\], then \[\alpha + \beta\] is equal to
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
Match each item given under column C1 to its correct answer given under column C2.
| C1 | C2 |
| (a) `(1 - cosx)/sinx` | (i) `cot^2 x/2` |
| (b) `(1 + cosx)/(1 - cosx)` | (ii) `cot x/2` |
| (c) `(1 + cosx)/sinx` | (iii) `|cos x + sin x|` |
| (d) `sqrt(1 + sin 2x)` | (iv) `tan x/2` |
If cotθ + tanθ = 2cosecθ, then find the general value of θ.
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`]
If sinθ + cosθ = 1, then the value of sin2θ is equal to ______.
Given x > 0, the values of f(x) = `-3cos sqrt(3 + x + x^2)` lie in the interval ______.
State whether the statement is True or False? Also give justification.
If tanA = `(1 - cos B)/sinB`, then tan2A = tanB
State whether the statement is True or False? Also give justification.
If tanθ + tan2θ + `sqrt(3)` tanθ tan2θ = `sqrt(3)`, then θ = `("n"pi)/3 + pi/9`
