Asked by Anonymous
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
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
Answers
Answered by
Houdini
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
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
Answered by
Reiny
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!!!
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