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# Trigonometric question?

If sinx+cosx = -√2

What is sin^3 x+ cos^5 x=?

### 3 Answers

- 1 month ago
Let's pretend we don't know that x must be 5pi/4 radians

sin(x) + cos(x) = -sqrt(2)

(sin(x) + cos(x))^2 = (-sqrt(2))^2

sin(x)^2 + 2sin(x)cos(x) + cos(x)^2 = 2

1 + sin(2x) = 2

sin(2x) = 1

2x = arcsin(1)

2x = pi/2 + 2pi * k

x = pi/4 + pi * k

k is an integer.

Just using basic trig identities, namely sin(x)^2 + cos(x)^2 = 1 and 2sin(x)cos(x) = sin(2x), we have possible values for x

From 0 to 2pi we have 2 possible values: pi/4 and pi/4 + pi, or 5pi/4

sin(pi/4) + cos(pi/4) = sqrt(2)/2 + sqrt(2)/2 = sqrt(2)

So that's not it

sin(5pi/4) + cos(5pi/4) = -sqrt(2)/2 - sqrt(2)/2 = -sqrt(2)

Bingo

sin(x)^3 + cos(x)^5 =>

sin(5pi/4)^3 + cos(5pi/4)^5 =>

(-sqrt(2)/2)^3 + (-sqrt(2)/2)^5 =>

(-sqrt(2)/2)^3 * (1 + (-sqrt(2)/2)^2) =>

(-2 * sqrt(2) / 8) * (1 + 2/4) =>

(-sqrt(2)/4) * (1 + 1/2) =>

(-sqrt(2)/4) * (3/2) =>

-3 * sqrt(2) / 8

- 1 month ago
x = 225 degrees (subtract cos x from both sides, square both sides, use sin^2 = 1 - cos^2, solve quadradic equation).

sin x = -sqrt(2)/2, cos x = -sqrt(2)/2

(-sqrt2/2)^3 = - 2sqrt2/8 = -sqrt(2)/4

(-sqrt2/2)^5 = - 4sqrt2/32 = -sqrt(2)/8

answer = - 3 sqrt(2)/8

- hayharbrLv 71 month ago
If sinx+cosx = -√2 then sin x = cos x = -√2 / 2

so x could be 5π/4

not that it matters...

(-√2/2)^3 = -√8 / 8 = -2√2 / 8 = -√2 / 4

(-√2/2)^5 = -√32 / 32 = -4√2 / 32 = -√2 / 8 so add those

imo anyway