Trigonometry question: sine and cosine

[astonky]

Hi,

For my homework, I need to solve the following question:

Given that cos a = 4/5 and cos b = 7/25, find the possible values of cos (a+b).

What I have done so far is this:

 

Using compound angle identites: cos (a+b) = cos a cos b - sin a sin b

Therefore: cos (a+b) = (4/5)*(7/25) - sin a sin b

I now used a calculator to find the values of cos a and cos b to find the values of sin a and sin b. Thus: sin a = 3/5 and sin b = 24/25

Then substituted back into compound angle formula: cos (a+b) = (4/5)*(7/25) - (3/5)*(24/25)

Simplifying gives me 28/125 - 72/125, and thus

cos (a+b)= -44/125.

Then I found the angle cos (a+b) using a calculator: 111°

And used that to find the second possible angle: 180°-111°= 69°

Then I did took cos (69°) to get the second value of cos (a+b), which according to the calculator is 0.351.

But according to my textbook answers, the second value of cos (a+b) is 4/5. Can you help explain where I made a mistake, and even more is there a shorter way to solve this type of question?

 

Thanks for you help.

[Vioh]

Some small tips:

1. It's good (and even encouraged) to use the calculator as much as possible during the exam. However, for the purpose of practice (especially for HL students), I think it's best to use the calculator as little as possible. The reason is that the calculator might give you a false sense of confidence, leading you to think that you understand stuff, while in reality you don't. For this question, you should only use the calculator for tedious stuff like multiplication (25^2 and 25*5, or something like that), but you shouldn't use it for solving for sin(a) and sin(b).
2. Unless you are dealing with triangles which is not in this case, avoid using degrees all together. Always use radians, because you'll less likely make a mistake.

Regarding the problem, you are close. Your first calculation (that yields -44/125) is done correctly. However, you've missed the other solution entirely. The part where you calculate stuff using degrees is completely necessary (and even wrong). I don't understand why you would calculate 180 - 111. What's the point of that?

The reason why you didn't get the 2nd solution is because there are other possible values for sin(a) and sin(b) that you've missed. To calculate these 2 expressions, you need to use the Pythagorean trigonometric identity:

1 = sin²(a) + cos²(a) = sin²(a) + (4/5)² ==> sin(a) = ±3/5
1 = sin²(b) + cos²(b) = sin²(b) + (7/25)² ==> sin(b) = ±24/25
Hence, sin(a)sin(b) = ±72/125

Substitute this into the compound angle identity, you have cos(a+b) = 21/125 ± 72/125 = {-44/125, 100/125} = {-44/125, 4/5}. Hope this clear up your confusion.

 

[kw0573]

Positive cosine is in quadrant I and IV, where sin is positive in I and negative in IV. It can be show that whether both angles in I or both in IV, the cos (a + b) is same value. One angle in each of the two quadrants results in the other cos (a + b) value. Assume some positive value of a and b from 0 to pi/2. Because cos(b) = cos(-b), you use either a and b for cos (a+b) or a and -b in which case you evaluate cos (a-b). As you can see, the two sums of angles dont add to 180, but rather is averaged to a.Hence your 180- angle method is erroneous.Hope that helps!

[astonky]

Thanks a lot, kw0573 and Vioh. This forums is a real life saver for math hl