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Given that sin A=3/7, cos B=-2/5, and both AA and B are in quadrant II, find cos (A-B). Simplify to a single value and leave it in the form of a rational number.

HELP PLEASE Given that sin A37 cos B25 and both AA and B are in quadrant II find cos AB Simplify to a single value and leave it in the form of a rational number class=

Respuesta :

LammettHash

First, recall that

cos(A - B) = cos(A) cos(B) + sin(A) sin(B)

so you just need to find cos(A) and sin(B).

Since both A and B end in the second quadrant, you know that

• cos(A) and cos(B) are both negative

• sin(A) and sin(B) are both positive

Then from the Pythagorean identity, you get

cos²(A) + sin²(A) = 1   ==>   cos(A) = -√(1 - sin²(A)) = -2√10/7

cos²(B) + sin²(B) = 1   ==>   sin(B) = +√(1 - cos²(B)) = √21/5

You'll end up with

cos(A - B) = (-2√10/7) (-2/5) + (3/7) (√21/5)

… = (4√10 + 3√21)/35

(which makes the last sentence in the question kind of confusing, because this expression doesn't get much simpler and it's certainly not a rational number)

The value of cos(A - B) is approximately 23/25

Given that A and B are in the second quadrant, we have

  • sin A = 3/7
  • cos B = -2/5

To find cos(A - B), we have to use trigonometric functions

cos(A - B) = cosAcosB + sinAsinB ...equation(i)

but

cos A

[tex]cos^2A + sin^2A =1 \\cos^2A = 1 - sin^2A\\cos^2A = 1 - (\frac{3}{7})^2 = 1 - \frac{9}{49}= cosA= -\frac{2\sqrt{5} }{7}[/tex]

Having the value of cos A, let's solve for cosB

Cos B

cos B = -2/5

[tex]sin^2B = 1-cos^2B\\sin^2B = 1-(-\frac{2}{5})^2= 1-\frac{4}{25}\\sinB = \sqrt{\frac{21}{25} }=\frac{\sqrt{21} }{5}[/tex]

cos(A-B)

substituting the values if sinA, cosA, sinB, cosB into equation(i) above;

[tex]cos(A-B)=cosAcosB+sinAsinB\\cos(A-B)=(-\frac{2\sqrt{5} }{7})(-\frac{2}{5})+(\frac{3}{7})(\frac{\sqrt{21} }{5})\\cos(A-B)=\frac{3\sqrt{21}+4\sqrt{5} }{35} \\cos(A-B) = 23/35[/tex]

The value of cos(A-B) is given above

Learn more on trigonometric functions here;

https://brainly.com/question/4326804

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