Question

sin0 + cos0 = 7/6
what is sin0 - cos0 = ?
0 represents theta

Answer 1

\[2(\pi n+\arctan(\frac{1}{13}(6-\sqrt{23})))\]

Answer 2

\[n\in \mathbb{Z}\]

Answer 3

\[sin\theta + cos\theta = \frac{7}{6}\]\[(sin\theta + cos\theta)^2 = (\frac{7}{6})^2\]\[sin^2\theta + cos^2\theta+2sin\theta cos\theta= \frac{49}{36}\]\[1+2sin\theta cos\theta= \frac{49}{36}\]\[2sin\theta cos\theta= \frac{13}{36}\]

\[sin\theta - cos\theta\]\[=\sqrt{(sin\theta - cos\theta)^2}\]\[=\sqrt{(sin^2\theta-2sin\theta\cos\theta+\cos^2\theta)}\]You know what \(sin^2\theta + cos^2\theta\) and \(2sin\theta\cos\theta\) are, so you can find the answer.

Answer 4

@Hyun11 Can you explain how you get your answer?

Answer 5

^ http://wolframalpha.com

Answer 6

sin0 + cos0 = 7/6
let sin0 - cos0 = x

square both sides of both eqns and add them, you'll get your answer in the very next step.

Answer 7

how if you are asked \[(\sin \theta)^2 - (\cos \theta)^2\] ? @Callisto

Answer 8

Do you know how to solve for sin(theta) and cos(theta) given the two equations above?

Answer 9

a+b=7/6
a-b=(callisto basically gave you this answer)
Solve the system of linear equations

Answer 10

Plug into a^2-b^2 and you got it

Answer 11

thx, that helped a lot
@myininaya

Answer 12

np :)