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###### Need help with another AL Maths Question

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2 months ago

https://www.quora.com/profile/Bravewarrior/Need-help-with-part-a-of-this-question-1

Here is the question and solution. I'm confused as to how in part a they suddenly got 3sin theta. Any help would be appreciated!

Here is the question and solution. I'm confused as to how in part a they suddenly got 3sin theta. Any help would be appreciated!

Original post by pigeonwarrior

https://www.quora.com/profile/Bravewarrior/Need-help-with-part-a-of-this-question-1

Here is the question and solution. I'm confused as to how in part a they suddenly got 3sin theta. Any help would be appreciated!

Here is the question and solution. I'm confused as to how in part a they suddenly got 3sin theta. Any help would be appreciated!

By definition, cos is the COmplementary angle SIN so

cos(x) = sin(pi/2 - x)

or

cos(pi/2 - x) = sin(x)

Its one of the basic identities which drops out of a simple right triangle, which has two complementary angles.

Reply 2

2 months ago

Original post by mqb2766

By definition, cos is the COmplementary angle SIN so

cos(x) = sin(pi/2 - x)

or

cos(pi/2 - x) = sin(x)

Its one of the basic identities which drops out of a simple right triangle, which has two complementary angles.

cos(x) = sin(pi/2 - x)

or

cos(pi/2 - x) = sin(x)

Its one of the basic identities which drops out of a simple right triangle, which has two complementary angles.

Ohhh so as the cosine graph is just a shifted version of the sine graph, for this question we assume that cos(pi/2-x) is the same as sine? As it is so similar to the sine graph?

Original post by pigeonwarrior

Ohhh so as the cosine graph is just a shifted version of the sine graph, for this question we assume that cos(pi/2-x) is the same as sine? As it is so similar to the sine graph?

The previous (two) identity is how cos is defined. So if you have

A+B=90

Then

cos(A)=sin(B)

and sin(A)=cos(B) and they should be clear if you sketch a right triangle and think about which side ratios they refer to. So as you say the cos and sin curves are identical apart from a shift by pi/2.

Edit - if you didbt spot it, you could expand cos(pi/2-theta) using the usual angle addition identity and get the above the long way.

(edited 2 months ago)

Reply 4

2 months ago

Original post by mqb2766

The previous (two) identity is how cos is defined. So if you have

A+B=90

Then

cos(A)=sin(B)

and sin(A)=cos(B) and they should be clear if you sketch a right triangle and think about which side ratios they refer to. So as you say the cos and sin curves are identical apart from a shift by pi/2.

Edit - if you didbt spot it, you could expand cos(pi/2-theta) using the usual angle addition identity and get the above the long way.

A+B=90

Then

cos(A)=sin(B)

and sin(A)=cos(B) and they should be clear if you sketch a right triangle and think about which side ratios they refer to. So as you say the cos and sin curves are identical apart from a shift by pi/2.

Edit - if you didbt spot it, you could expand cos(pi/2-theta) using the usual angle addition identity and get the above the long way.

Ohhh ok thank you so so much!!!

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