# Answer on Trigonometry Question for Richard Johnson

Question #9824

Hi I am really struggling with this:

Re-write the expression 5.5 Cos t + 7.8 Sin t, in the form R sin (t+α).

Any help would be greatly appreciated as im pulling my hair out with this now!

Re-write the expression 5.5 Cos t + 7.8 Sin t, in the form R sin (t+α).

Any help would be greatly appreciated as im pulling my hair out with this now!

Expert's answer

5.5 Cos t + 7.8 Sin t=sqrt(5.5^2+7.8^2)*(Cos t *5.5/sqrt(5.5^2+7.8^2)+Sin t *7.8 /sqrt(5.5^2+7.8^2))

sqrt(5.5^2+7.8^2)= approx. 9.5

so we have

5.5 Cos t + 7.8 Sin t=9.5*(Cos t *5.5/9.5+Sin t *7.8 /9.5)

We can denote 5.5/9.5=sin(theta), 7.8 /9.5=cos(theta) because& 7.8^2+5.5^2=9.5^2 therefore sin(theta^2)+cos(theta)^2=1 so that's all right

now we can use formula fopr sine of the sum:

5.5 Cos t + 7.8 Sin t=9.5*(Cos t *5.5/9.5+Sin t *7.8 /9.5)=9.5 sin(t+theta) where theta is defined above

In your denotations it will be 5.5 Cos t + 7.8 Sin t=R sin(t+α) where R=9.5, sin α=5.5/9.5, cos α=7.8 /9.5

sqrt(5.5^2+7.8^2)= approx. 9.5

so we have

5.5 Cos t + 7.8 Sin t=9.5*(Cos t *5.5/9.5+Sin t *7.8 /9.5)

We can denote 5.5/9.5=sin(theta), 7.8 /9.5=cos(theta) because& 7.8^2+5.5^2=9.5^2 therefore sin(theta^2)+cos(theta)^2=1 so that's all right

now we can use formula fopr sine of the sum:

5.5 Cos t + 7.8 Sin t=9.5*(Cos t *5.5/9.5+Sin t *7.8 /9.5)=9.5 sin(t+theta) where theta is defined above

In your denotations it will be 5.5 Cos t + 7.8 Sin t=R sin(t+α) where R=9.5, sin α=5.5/9.5, cos α=7.8 /9.5

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