PLEASE HELP!!

Solve:
1. cos(arcsin(4/5))= ??
2. tan(arccos(4/5))= ??
3. cos(arctan(3))= ??
Note: Give exact answers!!

Let's first understand what this kind of question means.

I will do the 3rd one

cos(arctan(3))

arctan(3/1) means the angle Ø so that tanØ = 3/1
so draw a right angles triangle with angle Ø, opposite as 3 and adjacent as 1, ( remember tanØ = opp/adj )
By Pythagoras, the hypotenuse is √10

so cos(arctan(3))
= cos Ø
= 1/√10

Do the first 2 the same way.

I got them :). Thank you so much for explaining it really well, my textbook doesn't explain well at all, that's why I was struggling!! Thank you once again!!

You are welcome,

90% of success in math is understanding it

To solve these problems, we will use the trigonometric identities and properties. Let's go step by step:

1. cos(arcsin(4/5))
We can use the Pythagorean identity to solve this. Recall that the Pythagorean identity states: sin^2(theta) + cos^2(theta) = 1.
Since we have arcsin(4/5), let's assume sin(theta) = 4/5.
Using the identity, we can calculate cos(theta):
cos^2(theta) = 1 - sin^2(theta)
cos^2(theta) = 1 - (4/5)^2
cos^2(theta) = 1 - 16/25
cos^2(theta) = 9/25
Taking the square root of both sides, we get:
cos(theta) = +/- sqrt(9/25)
cos(theta) = +/- (3/5)
Since we are looking for the exact answer, let's take cos(theta) = 3/5.
Therefore, cos(arcsin(4/5)) = 3/5.

2. tan(arccos(4/5))
We can use the same approach as in the previous question. We will assume cos(theta) = 4/5.
Using the Pythagorean identity, we can calculate sin(theta):
sin^2(theta) = 1 - cos^2(theta)
sin^2(theta) = 1 - (4/5)^2
sin^2(theta) = 1 - 16/25
sin^2(theta) = 9/25
Taking the square root of both sides, we get:
sin(theta) = +/- sqrt(9/25)
sin(theta) = +/- (3/5)
Since we are looking for the exact answer, let's take sin(theta) = 3/5.
Now, use the tangent identity: tan(theta) = sin(theta) / cos(theta)
tan(theta) = (3/5) / (4/5)
tan(theta) = (3/5) * (5/4)
tan(theta) = 3/4
Therefore, tan(arccos(4/5)) = 3/4.

3. cos(arctan(3))
We will assume tan(theta) = 3.
Using the Pythagorean identity, we can calculate cos(theta):
cos^2(theta) = 1 / (1 + tan^2(theta))
cos^2(theta) = 1 / (1 + 3^2)
cos^2(theta) = 1 / (1 + 9)
cos^2(theta) = 1 / 10
Taking the square root of both sides, we get:
cos(theta) = +/- sqrt(1/10)
cos(theta) = +/- (1/√10)
Since we are looking for the exact answer, let's take cos(theta) = 1/√10.
Therefore, cos(arctan(3)) = 1/√10.