Advertisements
Advertisements
Question
If sinx + cosx = a, then |sinx – cosx| = ______.
Advertisements
Solution
Given that: sinx + cosx = a
(sinx + cosx)2 = a2
⇒ sin2x + cos2x + 2sinx cosx = a2
⇒ 1 + 2sinx cosx = a2
⇒ sinx cosx = `(a^2 - 1)/2` .......(i)
|sinx – cosx| = sin2x + cos2x – 2sinx cosx
= `1 - 2((a^2 - 1)/2)`
= 1 – (a2 – 1)
= 1 – a2 + 1
= 2 – a2
∴ |sinx – cosx| = `sqrt(2 - a^2)`
APPEARS IN
RELATED QUESTIONS
Find the value of: tan 15°
Prove the following: `cos (pi/4 xx x) cos (pi/4 - y) - sin (pi/4 - x)sin (pi/4 - y) = sin (x + y)`
Prove the following: `(tan(pi/4 + x))/(tan(pi/4 - x)) = ((1+ tan x)/(1- tan x))^2`
Prove the following:
`cos ((3pi)/ 2 + x ) cos(2pi + x) [cot ((3pi)/2 - x) + cot (2pi + x)]= 1`
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:
`(sin 5x + sin 3x)/(cos 5x + cos 3x) = tan 4x`
Prove the following:
`(cos 4x + cos 3x + cos 2x)/(sin 4x + sin 3x + sin 2x) = cot 3x`
Prove that: `(cos x + cos y)^2 + (sin x - sin y )^2 = 4 cos^2 (x + y)/2`
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)
Prove that:
\[\frac{7\pi}{12} + \cos\frac{\pi}{12} = \sin\frac{5\pi}{12} - \sin\frac{\pi}{12}\]
Prove that
If tan A = x tan B, prove that
\[\frac{\sin \left( A - B \right)}{\sin \left( A + B \right)} = \frac{x - 1}{x + 1}\]
Find the maximum and minimum values of each of the following trigonometrical expression:
12 cos x + 5 sin x + 4
Find the maximum and minimum values of each of the following trigonometrical expression:
\[5 \cos x + 3 \sin \left( \frac{\pi}{6} - x \right) + 4\]
If 12 sin x − 9sin2 x attains its maximum value at x = α, then write the value of sin α.
If tan (A + B) = p and tan (A − B) = q, then write the value of tan 2B.
If \[\frac{\cos \left( x - y \right)}{\cos \left( x + y \right)} = \frac{m}{n}\] then write the value of tan x tan y.
If tan \[\alpha = \frac{1}{1 + 2^{- x}}\] and \[\tan \beta = \frac{1}{1 + 2^{x + 1}}\] then write the value of α + β lying in the interval \[\left( 0, \frac{\pi}{2} \right)\]
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 in ∆ABC, tan A + tan B + tan C = 6, then cot A cot B cot C =
tan 3A − tan 2A − tan A =
If sin (π cos x) = cos (π sin x), then sin 2x = ______.
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
Express the following as the sum or difference of sines and cosines:
2 cos 3x sin 2xa
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 `(sin(x + y))/(sin(x - y)) = (a + b)/(a - b)`, then show that `tanx/tany = a/b` [Hint: Use Componendo and Dividendo].
Find the most general value of θ satisfying the equation tan θ = –1 and cos θ = `1/sqrt(2)`.
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 tan θ = 3 and θ lies in third quadrant, then the value of sin θ ______.
If tanθ = `a/b`, then bcos2θ + asin2θ 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 cosecx = 1 + cotx then x = 2nπ, 2nπ + `pi/2`
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`
