cos(2 tan^-1 1/3)
cos(sin^-1 1/5 +cos^-1 1/3)
For this one I got the answer to be sqrt 24/15 -sqrt 8/15, is that right?
I have no clue how to start the first one
In the future, please provide more information, it's nearly imposible to figure out what you're doing.
Are you simplifying? It's difficult to tell.
You may find it advantageous to know that:
Sin^2+cos^2=1
thus
1-cos^2=sin^2
1-sin^2=cos^2
and tan= Sin/cos
for the first one ...
tan^-1 (1/3) represents the angle Ø so that
tanØ = 1/3
construct a triangle in standard position in quadrant I
with opposite 1 and adjacent 3. (recall tanØ = opp/adj)
that makes the hypotenuse equal to √10
and sinØ=1/√10 and cosØ = 3/√10
so cos(2tan^-1 (1/3)
= cos 2Ø, we now know everything about Ø
but cos 2Ø = cos^2Ø - sin^2Ø
= 9/10 - 1/10 = 8/10 = 4/5
for the second,
let A = sin^-1 (1/5) and
let B = cos^-1 (1/3)
again draw triangles
for the angle A triangle, opp = 1, hyp = 5, then adj = √24
so sin A = 1/5, and cosA = √24/5
for the angle B triangle, adj = 1, hyp = 3, then opp = √8
so sinB = √8/3 and cosB = 1/3
so we really want
cos(A + B)
which expands to
cosAcosB - sinAsinB
= (√24/5)(1/3 - (1/5)(√8/3)
= (√24 - √8/15
= (2√6 - 2√2)/15
Very good, you had that!!!
To solve the first expression, cos(2 tan⁻¹(1/3)), we can start by using the identity:
tan(2θ) = 2tan(θ) / (1 - tan²(θ))
In this case, let's set θ = tan⁻¹(1/3), which means tan(θ) = 1/3. We substitute these values into the identity:
tan(2(tan⁻¹(1/3))) = 2(1/3) / (1 - (1/3)²)
Simplifying this expression gives us:
tan(2(tan⁻¹(1/3))) = 2/3 / (1 - 1/9)
= 2/3 / (8/9)
= 2/3 * 9/8
= 18/24
= 3/4
Now, since we have the value of tan(2(tan⁻¹(1/3))), we can use the double-angle formula for cosine:
cos(2(tan⁻¹(1/3))) = cos²(tan⁻¹(1/3)) − sin²(tan⁻¹(1/3))
= (1 - sin²(tan⁻¹(1/3))) / (1 + tan²(tan⁻¹(1/3)))
= (1 - (1/3)²) / (1 + (1/3)²)
= (1 - 1/9) / (1 + 1/9)
= 8/9
Therefore, cos(2 tan⁻¹(1/3)) = 8/9.
Now, let's move on to the second expression: cos(sin⁻¹(1/5) + cos⁻¹(1/3)).
To solve this expression, we can use the Pythagorean identity sin²(θ) + cos²(θ) = 1.
Let's consider the angles involved:
sin⁻¹(1/5) and cos⁻¹(1/3)
We can find the corresponding sides of a right-angled triangle as follows:
For sin⁻¹(1/5): opposite side = 1, hypotenuse = 5, adjacent side = √(5² - 1²) = 2√6.
For cos⁻¹(1/3): adjacent side = 1, hypotenuse = 3, opposite side = √(3² - 1²) = 2√2.
To find the sum of these angles, we can form a triangle using these two angles, where the adjacent side of sin⁻¹(1/5) is equal to the opposite side of cos⁻¹(1/3) (i.e., 2√6 = 2√2).
Now, we can use the Pythagorean identity to find the hypotenuse of the triangle:
hypotenuse = √((1)² + (2√6 + 2√2)²)
= √(1 + (2√6 + 2√2)²)
= √(1 + (2√6)² + (2√2)² + 2 * 2√6 * 2√2)
= √(1 + 4 * 6 + 4 * 2 + 8√12)
= √(1 + 24 + 8 + 8√12)
= √(33 + 8√12)
Finally, substituting the values into the expression:
cos(sin⁻¹(1/5) + cos⁻¹(1/3)) = cos(θ) = adjacent side / hypotenuse
= (2√2) / (√(33 + 8√12))
To simplify this expression further, rationalize the denominator by multiplying both the numerator and denominator by the conjugate of the denominator:
cos(sin⁻¹(1/5) + cos⁻¹(1/3)) = cos(θ) = (2√2 * (√(33 + 8√12))) / ((√(33 + 8√12)) * (√(33 + 8√12)))
= (2√2 * (√(33 + 8√12))) / (33 + 8√12)
At this point, it appears that you made an error in simplifying the expression. The correct answer should be:
cos(sin⁻¹(1/5) + cos⁻¹(1/3)) = (2√2 * (√(33 + 8√12))) / (33 + 8√12)
I hope this clarifies how to solve these two expressions! Let me know if you have any further questions.