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###### Help With A level Trig

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2 months ago

how do I find the values of theta for the equation: sin(60∘+θ)=cosθ

for 0∘≤θ≤360∘

Edit: I have got the lowest value which is 15, I am just unsure on how to work out the 2nd value

for 0∘≤θ≤360∘

Edit: I have got the lowest value which is 15, I am just unsure on how to work out the 2nd value

(edited 2 months ago)

Original post by max33456789

how do I find the values of theta for the equation: sin(60∘+θ)=cosθ

for 0∘≤θ≤360∘

Edit: I have got the lowest value which is 15, I am just unsure on how to work out the 2nd value

for 0∘≤θ≤360∘

Edit: I have got the lowest value which is 15, I am just unsure on how to work out the 2nd value

You could use the complementary identity for sin or cos.

Reply 2

2 months ago

Original post by mqb2766

You could use the complementary identity for sin or cos.

I used cos theta = sin (90 - theta)

I am unsure how to calculate the other value using it though

Original post by max33456789

I used cos theta = sin (90 - theta)

I am unsure how to calculate the other value using it though

I am unsure how to calculate the other value using it though

Thats what I meant. So when you do arcsin, what are the multiple values? If necessary, think of cast or sin curves.

Reply 4

2 months ago

Original post by mqb2766

Thats what I meant. So when you do arcsin, what are the multiple values? If necessary, think of cast or sin curves.

When I done it I done:

sin(60+theta) = sin(90-theta)

60+theta = 90-theta

Then solved to get theta = 15

How do I do the arcsin in this case?

Edit: Do I do sin 15 and then plot against a sin graph to then see where it intersects to give me my 2nd value?

(edited 2 months ago)

Original post by max33456789

When I done it I done:

sin(60+theta) = sin(90-theta)

60+theta = 90-theta

Then solved to get theta = 15

How do I do the arcsin in this case?

sin(60+theta) = sin(90-theta)

60+theta = 90-theta

Then solved to get theta = 15

How do I do the arcsin in this case?

When you go from line 1 to line 2 (bold) you take inverse sin or arcsin. sin has a restricted domain to make it invertible so when you do the inverse its got multiple solutions. As an example

sin(x) = sin(90-x)

is obviously satisfied by x=45 so sin(x) = 1/sqrt(2) but also by x=225

https://www.desmos.com/calculator/bybfbtk8ci

when sin(x)=-1/sqrt(x).

arcsin is has a range -90 to 90 and you have to explcitly think about solutions outside that domain.

(edited 2 months ago)

Reply 6

2 months ago

It would just be easier to use addition formulae.

Reply 7

2 months ago

Original post by mqb2766

When you go from line 1 to line 2 (bold) you take inverse sin or arcsin. sin has a restricted domain to make it invertible so when you do the inverse its got multiple solutions. As an example

sin(x) = sin(90-x)

is obviously satisfied by x=45 so sin(x) = 1/sqrt(2) but also by x=225

https://www.desmos.com/calculator/bybfbtk8ci

when sin(x)=-1/sqrt(x).

arcsin is has a range -90 to 90 and you have to explcitly think about solutions outside that domain.

sin(x) = sin(90-x)

is obviously satisfied by x=45 so sin(x) = 1/sqrt(2) but also by x=225

https://www.desmos.com/calculator/bybfbtk8ci

when sin(x)=-1/sqrt(x).

arcsin is has a range -90 to 90 and you have to explcitly think about solutions outside that domain.

Thank you I have got the right answer now. I had to add 180 each time, which makes sense now I look at the ATSC diagram

Original post by max33456789

Thank you I have got the right answer now. I had to add 180 each time, which makes sense now I look at the ATSC diagram

Thats correct, solutions occur every +/-k*180. Its down to the +/- solutions rather than the usual positive or negative symmetry in the cast diagram. As noted above you could do the angle addition identity and that would come out as tan() = ... and obv tan solutions repeat every 180.

(edited 2 months ago)

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