If sinA=4/3 find sin2A;cos2A

Answer 1

First, if sinA = 4/3, because 4/3 > 1, there is no ∠A ⇒ No solution.

However, I do want to show you how you would do these problems regardless: So here is a similar example:

Example: Suppose sinA = 4/5, where A is in Q2. Evaluate sin(2A) and cos(2A)

*Caution: Whatever you do, do not think of this as sin(2A) = 2(4/5) = 8/5, since 2A is just twice as much as A. It never works out that way!!

Using your double angle identities:

sin(2A) = 2sinAcosA............(1)

cos(2A) = cos²A - sin²A.......(2)

= 2cos²A - 1............(3)

= 1 - 2sin²A.............(4)

So to find sin(2A), based on the formula (1) above, we'll need to figure out the value of cosA. In this case since ∠A is in Q2, the value of cosA is negative. To figure out what cosA is, use your pythagorean identity:

cosA = √(1 - sin²A)

= √(1 - (4/5)²)

= √(1 - 16/25)

= √(9/25)

= 3/5

Since cosA is negative (since we're in Q2), cosA = -3/5

Thus, by (1) sin(2A) = 2sinAcosA

= 2(4/5)(-3/5)

= -24/25

As for cos(2A), you can use any of the formulas from (2) - (4) and get the same answer.

If you choose to use (2) If you choose to use (3) If you choose to use (4)

cos(2A) = cos²A - sin²A cos(2A) = 2cos²A - 1 cos(2A) = 1 - 2sin²A

= (-3/5)² - (4/5)² = 2(-3/5)² - 1 = 1 - 2(4/5)²

= 9/25 - 16/25 = 2(9/25) - 1 = 1 - 2(16/25)

= -7/25 = 18/25 - 1 = 1 - 32/25

= -7/25 = -7/25

As you can see, it really doesn't matter which of the three formulas you choose to use, you'll always end up with the same answer for cos(2A). Just pick your favorite one, and use it.

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One last note: Here's another way to see why sinA = 4/3 cannot produce an actual answer:

Since you need to find cosA

cosA = √(1 - sin²A)

= √(1 - (4/3)²)

= √(1 - 16/9)

= √(-7/9) = imaginary ⇒ No solution!

In general: The values of cosA and sinA should only be BETWEEN -1 and 1 (they may also include -1 and 1 itself) but they should never exceed them. So if someone gives you cosA = 5/3, since 5/3 > 1 ⇒ No solution

Hope this helps

Mr. K