sin theta and cos phi equal when theta greater than equal to zero or 90 greater than equal to phi
Answers
Answer:
We want to prove that the sine of an angle equals the cosine of its complement.
\sin(\theta) = \cos(90^\circ-\theta)sin(θ)=cos(90
∘
−θ)sine, left parenthesis, theta, right parenthesis, equals, cosine, left parenthesis, 90, degrees, minus, theta, right parenthesis
I'm skeptical. Please show me an example.
Let's start with a right triangle. Notice how the acute angles are complementary, sum to 90^\circ
∘
degrees.
Help! Please break this down for me.
Now here's the cool part. See how the sine of one acute angle
describes the \blueD{\text{exact same ratio}}exact same ratiostart color #11accd, start text, e, x, a, c, t, space, s, a, m, e, space, r, a, t, i, o, end text, end color #11accd as the cosine of the other acute angle?
Incredible! Both functions, \sin(\theta)sin(θ)sine, left parenthesis, theta, right parenthesis and \cos(90^\circ-\theta)cos(90
∘
−θ)cosine, left parenthesis, 90, degrees, minus, theta, right parenthesis, give the exact same side ratio in a right triangle.
And we're done! We've shown that \sin(\theta) = \cos(90^\circ-\theta)sin(θ)=cos(90
∘
−θ)sine, left parenthesis, theta, right parenthesis, equals, cosine, left parenthesis, 90, degrees, minus, theta, right parenthesis.
In other words, the sine of an angle equals the cosine of its complement.
Well, technically we've only shown this for angles between 0^\circ
∘
degrees and 90^\circ
∘
degrees. To make our proof work for all angles, we'd need to move beyond right triangle trigonometry into the world of unit circle trigonometry, but that's a task for another time.