Question

How to solve for cos427, sin1069 and cos61?

Not the decimal values, but using the values on the unit circle (e.g. √3/2, 1/2, √2/2)

Answer 1

First I should ask is the 427 in cos(427) degrees or radians? Same with 1069 and 61.

Answer 2

@aum they're all in degrees

Answer 3

The sine and cosine functions are periodic and they repeat themselves every 360 degrees or 2pi radians. Assuming the numbers are in degrees, we can keep subtracting as many multiples of 360 degrees from it and the value of the function will remain the same.
cos(427) = cos(427-360) = cos(67)
1069 - 360 = 709
709 - 360 = 349. The reference angle of 349 degrees is 360-349 = 11 degrees.
And since 349 is in the 4th quadrant where sine is negative, sin(349) = -sin(11).

Answer 4

The unit circle will give us sine, cosine values for standard angles such as 0, 30, 45, 60, 90 degrees, etc. but not for 67 degrees or 11 degrees. But we can find sin/cos for in between angles such as 15 degrees or 75 degrees by using addition and subtraction formula. For example sin(15) = sin(45-30) = sin(45)cos(30)-cos(45)sin(30).

But not for angles such as 67 degrees or 11 degrees.

Answer 5

@aum so the only possible answers are really just in the decimal form?

Answer 6

I think so or we can give an estimate or approximate value. For example. 61 degrees is so close to 60 degrees we can say cos(61) is approx cos(60), etc. Is this a problem form a textbook or from a teacher and for what class is it?

Answer 7

Well?

Answer 8

@aum sorry I had to step out for a while. Thanks for your answers I think these are helpful enough. Appreciate it

Answer 9

Now looking at the other questions you posted AFTER this question, I am getting the impression you have not posted the FULL question here. Probably you were not asked to find cos427, sin1069 and cos61 separately but they occur together in one question that involves trigonometric angle addition or subtraction which will change the angles to some standard angle. If you post the ENTIRE question and tag me I can help you solve it.

Answer 10

@aum yes you are right. I did that because finding those values was the reason I couldn't solve the entire problem. Here is one of them:

1. As a function of theta, where 0<theta<90 degrees), cos427 degrees=??

Still trying to look for the rest thanks so much for all your help

Answer 11

"As a function of theta, where 0< theta < 90 degrees, cos427 degrees = ?"

All they want you to do here is to reduce cos(427) to cos(theta) where theta is in the first quadrant. They don't want you to evaluate it, or use values from the unit circle.

Since cosine is a periodic function, with period 2pi radians or 360 degrees, you can subtract as many 360 degrees from 427 until the angle is in the first quadrant.

cos(427) = cos(427-360) = cos(67).
theta = 67 degrees satisfies the condition 0<theta<90 degrees.
So the answer is cos(67). That is it.

Answer 12

Okay got it. How about in this case where there is a cos50? √3/2cos50-1/2sin50 @aum