Solve this trigonometric thing??!!?!?

Question

Okay, I know that it's using the cofunction identities and the even/odd identities, but other than that, I'm at a loss. Any help would be GREATLY appreciated and as always, I'll pick a best answer TODAY!

Find the exact value of cos (α + ß) if sin α = 3/5 and sin ß = 5/13, with α in Quadrant II and ß in Quadrant I.

Now I know that that means that the cos of α will be negative and the cos of ß will be positive, and I also know that this involves the cofunction identities somehow, but it's acting for the EXACT value, so I can't use the inverse of the sin values of α and ß. So how would I go about solving this? Thank you so much!!!

P.S. What kind of class (i.e. what year in high school math) would you find this in????

Ooh! Something just dawned on me! Do you have to use the Pythagorean theorem to find the cosines?!

Answer 1

"Ooh! Something just dawned on me! Do you have to use the Pythagorean theorem to find the cosines?!"

Yep, anytime you're giving something like sin(x) = 3/5. It's just a ratio of two sides of a right triangle. You can find the other side using Pythagorean Theorem and subsequently find any of the the trig functions by definition.


Give sin(α) = 3/5
sin(α) = opp/hyp
opp^2 + adj^2 = hyp^2
adj^2 = hyp^2 - opp^2
adj^2 = 25-9
adj^2 = 16
adj = 4
Though since α is in quadrant II this is actually -4
adj = -4

Same thing for sin(β) = 5/13
adj^2 = 13^2 - 5^2
adj^2 = 12
this is positive since β is quadrant I


You have all 3 sides.. and can figure out any trig function from it..

Therefore you can write cos(a+β) as
cos(α)cos(β) - sin(α)sin(β)
(-4/5) (12/13) - (3/5)(5/13)
-48/65 - 15/65
-63/65

To answer when this would be taught. Probably midway to late through a Trigonometry class or in a Precalculus class

Answer 2

cos (α + ß)
= cosα cosß - sinαsinß
= (-4/5)(12/13)-(3/5)(5/13)
= -63/65
----------
Ideas:
cosα is negative in the second quadrant.
-------------
This should be taught in pre-Calculus class. Actually, you can also use the identity: cos^2(x) + sin^2(x) = 1, to get cosine once you know sine.

Answer 3

cos (α + ß) =cosαcosβ-sinαsinβ
=cosacosB-(3/5)(5/13)


cosα=sin(pi/2+α)
=sin(pi/2+3/2)
=sin(3Pi/2)
= -1

cosß=sin(pi/2+ß)
=sin(pi/2+5/13)
=sin(16pi/26)
=sin(8pi/13)

plug those values in the first equation and solve

=-1sin(8pi/13)- 3/13
=-sin(8pi/13)-3/13

Answer 4

Do not despair...this is an easy problem once you recognize the idea behind the question.

cos(alpha + beta) = cos(alpha)*cos(beta) - sin(alpha)*sin(beta)

and

sin^2(alpha) + cos^2(alpha) = 1
sin^2(beta) + cos^2(beta) = 1

Knowns:
sin(alpha) = 3/5
sin(beta) = 5/13

You can use the Pythagoras problem, too. Trying this strategy first (draw the traingle to convince yourself),

Alpha is a triangle with sides 4, 3, 5 (4^2 + 3^2 = 5^2)
Beta is a triangle with sides 5, 12, 13 (5^2 + 12^2 = 13^2)

so, cos(alpha) = 4/5
cos(beta) = 12/13

cos(alpha+beta) = (4/5)*(12/13) - (3/5)*(5/13)
= 0.5077

Good luck!

Answer 5

3sin x - 4cos x = 2 3sin x = 2 + 4cos x squaring on both sides 9sin^2(x) = 4 + 16cos^2(x) + 16cos x 9( 1 - cos^2(x)) = 4 + 16cos^2(x) + 16cos x 9 - 9 cos^2(x) = 4 + 16cos^2(x) + 16cos x 25 cos^2(x) + 16 cos x - 5 = 0 cos x = [-16 ± √(256+500)] /50 = [-16 ± √756] /50 = [-16 ± 27.5] /50 = 0.23 or - 0.87 x = 76.7, 283.3 degrees (when cos x = 0.23 ( Note:cos x = -0.87 is extraneous)

Answer 6

cos (α + β)
= cos α cos β - sinα sin β
= (- 4 / 5) (12 / 13) - (3 / 5) (5 / 13)
= - 48 / 65 - 15 / 65
= - 63 / 65

16/17 year old pupils.

Answer 7

cos(a + b) = cos(a)cos(b) - sin(a)sin(b)

this is part of the sum and difference formulas
that is one of them
the others are: cos(a - b), sin(a + b), sin(a - b)

look them up here :
http://www.mathematicshelpcentral.com/lecture_notes/precalculus_trigonometry_folder/sum_and_difference_formulas.htm