#Cos(π/2-x) =sinx

[quantumstudy]

Q: Find the range of the function y = 3 sinx + 4 cos(x + π/3) + 7

Solution : $\displaystyle cos (x + \frac{\pi}{3}) = cos x cos\frac{\pi}{3} - sinx sin\frac{\pi}{3} $

$\displaystyle cos (x + \frac{\pi}{3}) = \frac{1}{2} cos x - \frac{\sqrt{3}}{2} sinx $

$\displaystyle y = 3 sinx + 4 (\frac{1}{2} cos x - \frac{\sqrt{3}}{2} sinx) + 7 $

$\displaystyle y = 3 sinx + 2 cos x - 2 \sqrt{3} sinx + 7 $

$\displaystyle y = ( 3 - 2 \sqrt{3} ) sinx + 2 cos x + 7 $

Put $\displaystyle 3 - 2\sqrt{3} = r cos \alpha$ ...(i)

and $ 2 = r sin\alpha $  ...(ii)

$\displaystyle y = r sinx . cos \alpha + r cos x . sin\alpha + 7 $

$\displaystyle y = r sin(x + \alpha) + 7 $

Range of sin(x+α) is -1 to 1

Range of y = -r + 7 , r + 7

Now by squaring & adding equation (i) & (ii) we can get the value of r .

[devphilamaths]

Cos(π/2-x) =sinx solve trigonometric identities | Cos(pi/2-theta) =sintheta solve 12th Class Math

Cos(π/2-x) =sinx

[worldpedias]

How To Integrate Sin^2 <3

Let’s start off with an integral that we should already be able to do. X function with respect to x is equal to sum of the negative cos.

Integral of sin^2(nx)/sin^2(x) from 0 to pi YouTube from www.youtube.com

How to integrate sin^2. Integrate 1/(cos(x)+2) from 0 to 2pi; For each of these, we simply use the fundamental of calculus, because we know their corresponding derivatives. The integration of sine inverse is of the form.

Here are some examples illustrating how to ask for an integral. , csc cot, sec tan, csc. ∫ cosxsin5xdx = ∫ u5du using the substitution u =sinx = 1 6 sin6x+c ∫ cos.

Z = 2 π x, then. These is special integral fresnel integral. We have to integrate of sin 2 x.

We intend to travel a simple path from 0 to x, but we end up with a smaller percentage instead. Our calculator allows you to check your solutions to calculus exercises. = s(u) plug in solved integrals:

D x = π 2 d z, and. S (z) = ∫ sin ⁡ (π z 2 2) d z (1) let. The integral of sin(x) multiplies our intended path length (from 0 to x) by a percentage.

We recall the pythagorean identity and rearrange it for cos 2 x. X function with respect to x is written in the following mathematical form in calculus. All common integration techniques and even special functions are supported.

[vaish19-blog]

Ad maths Trignometric identities

1. If none of the angles x, y and (x + y) is an odd multiple of π/2, then tan (x + y) = (tan x + tan y)/(1 - tan x tan y) Since none of the x, y and (x + y) is an odd multiple of π/2, it follows that cos x, cos y and cos (x + y) are non-zero. Now tan (x + y) = sin (x + y)/ cos (x + y)  wkt sin (x + y) = sin x cos y + cos x sin y cos (x + y) = cos x cos y - sin x sin y So tan (x + y) = [ sin x cos y + cos x sin y ]/[ cos x cos y - sin x sin y ] Dividing numerator and denominator by cos x cos y, we have tan (x + y) = [( sin x cos y + cos x sin y )/(coa x cos y)]/[( cos x cos y - sin x sin y )/(cos x cos y)] tan (x + y) = [(sin x/cos x) + (sin y/cos y)]/[1 - (sin x/cos x)(sin y/cis y)] so tan (x + y) = (tan x + tan y)/(1 - tan x tan y) 2. tan ( x – y) = (tan x - tan y)/(1 + tan x tan y) If we replace y by – y in Identity 1, we get tan (x – y) = tan [x + (– y)] tan ( x – y) = (tan x - tan y)/(1 + tan x tan y) 3. If none of the angles x, y and (x + y) is a multiple of π, then cot ( x + y) = (cot x cot y - 1)/(cot y + cot x) Since, none of the x, y and (x + y) is multiple of π, we find that sin x sin y and sin (x + y) are non-zero. Now, cot ( x + y) = cos (x + y)/sin (x + y) wkt sin (x + y) = sin x cos y + cos x sin y cos (x + y) = cos x cos y - sin x sin y so cot (x + y) = (cos x cos y - sin x sin y)/( sin x cos y + cos x sin y ) Dividing numerator and denominator by sin x sin y, we have cot (x + y) = [(cos x cos y - sin x sin y )/(sin x sin y)]/[( sin x cos y + cos x sin y )/(sin x sin y)] cot (x + y) = (cot x cot y - 1)/(cot y + cot x) 4. cot (x – y) = (cot x cot y + 1)/(cot y - cot x) If we replace y by –y in identity 3, we get the result 14. cos 2x = (cosx)^2 – (sinx)^2 = 2 (cosx)^2 – 1 = 1 – 2 (sin x)^2 = [1 - (tan x)^2]/[ 1 + (tan x)^2 ] We know that cos (x + y) = cos x cos y – sin x sin y Replacing y by x, we get cos 2x = (cosx)^2 – (sin x)^2 = (cosx)^2 – [1 – (cos x)^2] = 2 (cosx)^2 – 1 Again, cos 2x = (cosx)^2 – (sin x)^2 = 1 – (sin x)^2 – (sin x)^2 = 1 – 2 (sin x)^2 We have cos 2x = (cosx)^2 – (sin x)^2 = [ (cosx)^2 – (sin x)^2 ]/[ (cosx)^2 + (sin x)^2 ] Dividing numerator and denominator by (cosx)^2, we get cos 2x = [1 - (tan x)^2]/[ 1 + (tan x)^2 ] x is not equal to nπ + π/2 where n is an integer 15. sin 2x = 2 sinx cos x = 2 tanx/ (1 + (tan x)^2) We have sin (x + y) = sin x cos y + cos x sin y Replacing y by x, we get sin 2x = 2 sin x cos x. Again sin 2x = 2 sinx cos x / [ (cosx)^2 + (sin x)^2 ] Dividing each term by (cos x)^2, we get sin 2x = [( 2 sinx cos x )/ (cos x)^2 ]/ [{(cosx)^2 + (sin x)^2}/ (cos x)^2 ] sin 2x = 2 tanx/ (1 + (tan x)^2)

[vaish19-blog]

Ad maths Trignometric identities

1. cos( π/2 - x) = sin x wkt cos (x - y) = cos x cos y + sin x sin y replace x by π/2 and 'y' by 'x' in above identity we get cos (π/2 - x) = cos (π/2) cos x + sin (π/2) sin x cos (π/2) = 0 and sin (π/2) = 1 so, cos (π/2 - x) = sinx 2. sin( π/2 - x) = cos x From the above Sin (π/2 - x) = cos [π/2 - (π/2 - x)] so sin(π/2 - x) = sin x 3. sin (x + y) = sin x cos y + cos x sin y we can write sin (x + y) = cos [π/2 - (x + y)] = cos [(π/2 - x) - y] so cos (x - y) = cos x cos y + sin x sin y so sin (x + y) = cos [(π/2 - x) - y] = cos (π/2 - x) cos y + sin (π/2 - x) siny sin(x + y) = sin x cos y + cos x sin y 4. sin (x - y) = sin x cos y - cos x sin y In the above identity replace y by '-y'. 5. similarly cos(π/2 + x) = -sin x sin(π/2 + x) = cos x cos(π - x) = -cos x sin(π - x) = sin x cos(π + x) = -cos x sin(π + x) = -sin x cos(2π - x) = cos x sin(2π - x) = -sin x Similar results canbe obtained for tan x, cotx,sec x, and cosec x.