Complementary Angles - Circular functions

[jashaan]

I was wondering whether we need to memorise the complementary/supplementary angle rules and are there many questions related to it in the exam?

Thank you!

[Sine]

I was wondering whether we need to memorise the complementary/supplementary angle rules and are there many questions related to it in the exam?

Thank you!

It is definitely not worthwhile rote-learning all those angle rules but make sure you understand them and have the ability to derive them if necessary e.g. using the unit circle

A possible question would be something like evaluate sin(pi/2 - theta)
Assuming theta is between 0 and pi/2 (i.e. within the 1st quadrant)

The way I would solve this is picture pi/2 on a unit circle and visualise a deduction of theta from that. Thus, that angle is within the 1st quadrant. In the 1st quadrant sin/cos/tan are all positive.

So sin(pi/2 - theta) = cos(theta)


Another example would be solve cos(3pi/2 - theta)
thus this angle in the 3rd quadrant. Sin is negative here so.
cos(3pi/2 - theta) = -sin(theta)

[jashaan]

It is definitely not worthwhile rote-learning all those angle rules but make sure you understand them and have the ability to derive them if necessary e.g. using the unit circle

A possible question would be something like evaluate sin(pi/2 - theta)
Assuming theta is between 0 and pi/2 (i.e. within the 1st quadrant)

The way I would solve this is picture pi/2 on a unit circle and visualise a deduction of theta from that. Thus, that angle is within the 1st quadrant. In the 1st quadrant sin/cos/tan are all positive.

So sin(pi/2 - theta) = cos(theta)


Another example would be solve cos(3pi/2 - theta)
thus this angle in the 3rd quadrant. Sin is negative here so.
cos(3pi/2 - theta) = -sin(theta)

Thank you!

But why is sin(pi/2+theta) = cos theta, considering that is second quadrant and in that, cos is negative??
and also cos(pi/2 + theta) = -sin theta, considering that is first quadrant and sin is meant to be positive?

MOD EDIT: merged posts

[Sine]

But why is sin(pi/2+theta) = cos theta, considering that is second quadrant and in that, cos is negative??
and also cos(pi/2 + theta) = -sin theta, considering that is first quadrant and sin is meant to be positive?

For your second equation that is actually the 2nd quadrant but yes sin is positive there as well.

I might not have been clear in my first post but as a rule, you should consider what quadrant the original circular function is in. Not what circular function you are trying to equate it to.

E.g. cos(pi/2 + theta) is in the 2nd quadrant.  Thus you know cos(pi/2 + theta) is negative.

[RuiAce]

For your second equation that is actually the 2nd quadrant but yes sin is positive there as well.

I might not have been clear in my first post but as a rule, you should consider what quadrant the original circular function is in. Not what circular function you are trying to equate it to.

E.g. cos(pi/2 + theta) is in the 2nd quadrant.  Thus you know cos(pi/2 + theta) is negative.

Shouldn't you technically say that pi/2 + theta is in the 2nd quadrant, and hence cos(pi/2 + theta) is negative?

[Sine]

Shouldn't you technically say that pi/2 + theta is in the 2nd quadrant, and hence cos(pi/2 + theta) is negative?

YES 

EDIT: my bad

[jashaan]

YES 

EDIT: my bad

Hahaha, that's alright, but yes I understand what you meant Thank you!