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
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
2 answers
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!!!
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!!!
1 answer