What is the solution to cos[arcsin (-4/5) + arccos (12/13)]?

[dkotschessaa]

Homework Statement

Evaluate the trigonometric expression (several examples). Hint: In each case, sketch a right triangle. For example, in the first case, sketch a right triangle with a hypotenuse of length 5 and one leg of length 3. Let A be the angle opposite the side of length 3. Solve for the length of the third side, and then look at the sketch to find tanA.

First example (which I solved) tan(arcsin 3/5)

Second Example (I'm boggled ) cos[arcsin (-4/5) + arccos (12/13)]

Homework Equations

For the first example tan(arcsin 3/5)

The missing side of the triangle by the Pythagorean theorem turns out to be 4 ,so tanA= opposite (3) over adjacent (4) = 3/4. No problem.

For the second...

cos[arcsin (-4/5) + arccos (12/13)]

The Attempt at a Solution

Since I've been specifically asked to draw a right triangle, I go along with it - I draw one with a side of -4 and solve the missing side for 3. The second triangle I get a missing side of 5.

Of course I can use a calculator and calculate the angles, add them together and take the cosine, but that's not what's being asked. I've tried various means of seeing if we're talking about one triangle instead of two, by drawing one a hypotenuse of length 65 (multiplying the hypotenuse of each). The other two sides are 60 and -40. This doesn't help. I know the answer is 56/65 I don't know how to get to it.

I've even tried putting the hypotenuse of triangle A as the opposite side of triangle B but I just get some bizarre shape that doesn't tell me anything. I don't know any trig identities that involve adding a sin of one angle with the cosine of another, much less their inverses.

I'm basically not understanding conceptually what's being asked here. Even if I wasn't dealing with this -4 in the equation I wouldn't know how to add these two.

-DaveKA

 

[Mark44]

Homework Statement

Evaluate the trigonometric expression (several examples). Hint: In each case, sketch a right triangle. For example, in the first case, sketch a right triangle with a hypotenuse of length 5 and one leg of length 3. Let A be the angle opposite the side of length 3. Solve for the length of the third side, and then look at the sketch to find tanA.

First example (which I solved) tan(arcsin 3/5)

Second Example (I'm boggled ) cos[arcsin (-4/5) + arccos (12/13)]

Homework Equations

For the first example tan(arcsin 3/5)

The missing side of the triangle by the Pythagorean theorem turns out to be 4 ,so tanA= opposite (3) over adjacent (4) = 3/4. No problem.

For the second...

cos[arcsin (-4/5) + arccos (12/13)]

The Attempt at a Solution

Since I've been specifically asked to draw a right triangle, I go along with it - I draw one with a side of -4 and solve the missing side for 3. The second triangle I get a missing side of 5.

Of course I can use a calculator and calculate the angles, add them together and take the cosine, but that's not what's being asked. I've tried various means of seeing if we're talking about one triangle instead of two, by drawing one a hypotenuse of length 65 (multiplying the hypotenuse of each). The other two sides are 60 and -40. This doesn't help. I know the answer is 56/65 I don't know how to get to it.

I've even tried putting the hypotenuse of triangle A as the opposite side of triangle B but I just get some bizarre shape that doesn't tell me anything. I don't know any trig identities that involve adding a sin of one angle with the cosine of another, much less their inverses.

I'm basically not understanding conceptually what's being asked here. Even if I wasn't dealing with this -4 in the equation I wouldn't know how to add these two.

-DaveKA

Start by using the sum of angles identity for cosine: cos(A + B) = cosAcosB - sinAsinB, where A = arcsin(-4/5) and B = arccos(12/13). From there, you should be able to use right triangle trig to find the two remaining values.

 

[dkotschessaa]

Perfect, Mark. Thank you!