- #1

thatguythere

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## Homework Statement

Evaluate sin^-1(cos70°)

## Homework Equations

## The Attempt at a Solution

sin^-1(cos70°)=θ

sinθ=cos70°

sinθ=1/2

sinθ=∏/3

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- Thread starter thatguythere
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- #1

thatguythere

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Evaluate sin^-1(cos70°)

sin^-1(cos70°)=θ

sinθ=cos70°

sinθ=1/2

sinθ=∏/3

- #2

Tanya Sharma

- 1,540

- 135

## Homework Statement

Evaluate sin^-1(cos70°)

## Homework Equations

## The Attempt at a Solution

sin^-1(cos70°)=θ

sinθ=cos70°

sinθ=1/2

sinθ=∏/3

First use cos(90° - θ ) = sinθ and then sin

- #3

Mark44

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Also, cos(70°) ≠ 1/2 and there is no real number θ for which sin(θ) = ##\pi/3## > 1.

- #4

thatguythere

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Could you possibly explain the cos(90°-θ)=sinθ?

- #5

thatguythere

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Ah a trig identity I see.

So cos(90°-70°)=sinθ

cos(20°)=sinθ

0.94=sinθ

sin^-1(sinθ)=θ

sin^-1(0.94)=θ

0.94=θ

So cos(90°-70°)=sinθ

cos(20°)=sinθ

0.94=sinθ

sin^-1(sinθ)=θ

sin^-1(0.94)=θ

0.94=θ

Last edited:

- #6

Mark44

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No, cos(90°-70°) = cos(20°) = sin(70°).Ah a trig identity I see.

So cos(90°-70°)=sinθ

In general, cos(90° - θ) = sin(θ).

cos(20°)=sinθ

0.94=sinθ

sin^-1(sinθ)=θ

sin^-1(0.94)=θ

0.94=θ

Most of the above makes no sense, with each line having little or no relation to the preceding line. Your value for θ is incorrect, and you could easily check that it is incorrect by using a calculator.

- #7

thatguythere

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- #8

Mark44

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Because fAlright. So if cos(90°-x)=sinx, which in my case is 70°, I still do not understand why Tanya then encourages me to use sin^-1(sinθ)=θ.

There's really no need for you to use θ, here, and I think that it is confusing you.How is cos70° to be replaced by sinθ?

Not equivalent to - equal to.By the trig identities, sin(90°-x)=cosx, it is sin20° which is equivalent to cos70°.

cos(70°) = cos(90° - 20°) = sin(20°)

Make it simpler by getting rid of θ.Would my equation then be accurately represented by sin^-1(sin20°)=θ and then simply by using cancellation identities reduced to 20°=θ?

sin

- #9

thatguythere

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sin^-1(sin(20°))=20°=∏/9

- #10

Mark44

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Yes.sin^-1(sin(20°))=20°=∏/9

- #11

thatguythere

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- #12

Mark44

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Sure, you're welcome!

- #13

Tanya Sharma

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Thanks Mark for nicely guiding the OP.

- #14

Mark44

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That's what they pay me for!

Oh, wait - I'm an unpaid volunteer.

Oh, wait - I'm an unpaid volunteer.

- #15

Tanya Sharma

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That's what they pay me for!

Oh, wait - I'm an unpaid volunteer.

and an excellent mentor

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