Rewrite Trigonometric Expressions

[nycmathguy]

Section 5.4

I need help with 91 and 93.

 

[MathLover1]

write the trigonometric expressions in the following forms:
a) sqrt(a^2+b^2)*sin(B*theta +C)
b) sqrt(a^2+b^2)*cos(B*theta-C)

a* sin θ + b* cos θ ≡ R sin (θ + α) where R=sqrt(a^2+b^2)

91.
given: sin(θ )+cos(θ )

a)
sin(θ )+cos(θ ) =>a=1 and b=1, C=arctan (a/b)=arctan (1/1)=π/4

then r=sqrt(a^2+b^2)=sqrt(1^2+1^2)=sqrt(2)

sin(theta)+cos(theta) =sqrt(2) sin(θ + π/4)

b)
sin(θ )+cos(θ ) =sqrt(2)*cos(theta-π/4)

93. you can do it same way

 

[nycmathguy]

write the trigonometric expressions in the following forms:
a) sqrt(a^2+b^2)*sin(B*theta +C)
b) sqrt(a^2+b^2)*cos(B*theta-C)

a* sin θ + b* cos θ ≡ R sin (θ + α) where R=sqrt(a^2+b^2)

91.
given: sin(θ )+cos(θ )

a)
sin(θ )+cos(θ ) =>a=1 and b=1, C=arctan (a/b)=arctan (1/1)=π/4

then r=sqrt(a^2+b^2)=sqrt(1^2+1^2)=sqrt(2)

sin(theta)+cos(theta) =sqrt(2) sin(θ + π/4)

b)
sin(θ )+cos(θ ) =sqrt(2)*cos(theta-π/4)

93. you can do it same way

You said:

sin(θ )+cos(θ ) =>a=1 and b=1, C=arctan (a/b)=arctan (1/1)=π/4

Where did C come from?

Check out the link below.

Verify an Identity

 

[nycmathguy]

write the trigonometric expressions in the following forms:
a) sqrt(a^2+b^2)*sin(B*theta +C)
b) sqrt(a^2+b^2)*cos(B*theta-C)

a* sin θ + b* cos θ ≡ R sin (θ + α) where R=sqrt(a^2+b^2)

91.
given: sin(θ )+cos(θ )

a)
sin(θ )+cos(θ ) =>a=1 and b=1, C=arctan (a/b)=arctan (1/1)=π/4

then r=sqrt(a^2+b^2)=sqrt(1^2+1^2)=sqrt(2)

sin(theta)+cos(theta) =sqrt(2) sin(θ + π/4)

b)
sin(θ )+cos(θ ) =sqrt(2)*cos(theta-π/4)

93. you can do it same way

Moving forward, I will post questions for which Ron Larson and David Cohen (two different precalculus books) provide sample problems for. This makes a lot more sense than randomly selecting a bunch of problems for which there is nothing for me to use as a guide to answer similar problems. I am also wasting lots of time answering questions that are beyond my level of mathematics. So, skipping 93 here makes sense.