What the identity settles

You know sin A = 3/5 and nothing else. Can you find cos A and tan A?

Find a missing ratio

Yes, when A is acute. The identity gives cos A. Then sin A ÷ cos A gives tan A.

Take sin A = 3/5. Then sin²A = 9/25. The identity gives cos²A = 1 − 9/25. That is 25/25 − 9/25 = 16/25.

Two numbers square to 16/25: 4/5 and −4/5. Cos A is one length divided by another, so it is positive. So cos A = 4/5.

Then tan A = sin A ÷ cos A = 3/5 ÷ 4/5 = 3/5 × 5/4 = 3/4.

Two more identities

Divide every term of sin²A + cos²A = 1 by cos²A. Then sin²A ÷ cos²A is tan²A, because tan A = sin A ÷ cos A. Also cos²A ÷ cos²A is 1. So the left side becomes tan²A + 1.

The right side is 1 ÷ cos²A. That is sec²A, because sec A = 1 ÷ cos A.

Now divide every term by sin²A instead. Then cos²A ÷ sin²A is cot²A, because cot A = cos A ÷ sin A. Also sin²A ÷ sin²A is 1. So the left side becomes 1 + cot²A.

The right side is 1 ÷ sin²A. That is cosec²A, because cosec A = 1 ÷ sin A.

Rule: A is an acute angle. 1 + tan²A = sec²A and 1 + cot²A = cosec²A.

Try the lab to finish this lesson.