Question

convert the following expressions into the form rsin(theta+alpha). (-pi 13 years ago

Answer 1

is it rsin??

Answer 2

@sriramkumar... yes, it is \[rsin(\theta+\alpha). (-\pi<\alpha<\pi) \]\[Q: \sin(\theta)+\cos(\theta)\]

Answer 3

use angle sum identity
\[\sin(\theta + \alpha) = \sin \theta \cos \alpha + \sin \alpha \cos \theta\]
thus
\[r = \sin \alpha = \cos \alpha\]
\[\alpha = \frac{\pi}{4}\]

Answer 4

how to get α=π 4 ?

Answer 5

wait maybe i didn't think this through

Answer 6

ok
\[r = \frac{1}{\sin \pi/4} = \sqrt{2}\]

...by solving sin x = cos x, x = pi/4

Answer 7

\[\sin \theta + \cos \theta = \sqrt{2}\sin(\theta + \frac{\pi}{4})\]

Answer 8

sorry i dont get it the one you typed ...by solving sinx=cosx.
can you elaborate more:)

Answer 9

oh sorry
\[\sin \theta = \cos \theta\]
\[\rightarrow \tan \theta = 1\]
\[\rightarrow \theta = \tan^{-1} (1)\]
\[\theta = \frac{\pi}{4}\]

Answer 10

then, what if the question is \[\sin(\theta)+\sqrt{3} \cos (\theta)\]
how can i get alpha? @dumbcow

Answer 11

For Solving this Question, I would like to ask you that, do you have the detailed conception of : Basic Trigonometry

Answer 12

\[\sin \theta + \cos \theta = \sqrt{2}(1/\sqrt{2}\sin \theta +1/\sqrt{2}\cos \theta) = \sqrt{2}(\cos45\sin \theta + \sin45 \cos \theta) = \sqrt{2}(\sin(\theta + 45 )) => rsin(\theta+\alpha)\]

Answer 13

Understood?

Answer 14

@goformit100.. detailed basic of trigo? hmm i know until the addition formula such double angle, product to sum and sum to product?
any specific topics that i can refer to?

Answer 15

@sriramkumar.. i cannot view the full solution given..
sorry..

Answer 16

this is what you need
\[a\sin(x)+b\cos(x)=\sqrt{a^2+b^2}\sin(x+\theta)\]
where
\[\sin(\theta)=\frac{b}{\sqrt{a^2+b^2}}\] and \[\cos(\theta)=\frac{a}{\sqrt{a^2+b^2}}\]

Answer 17

the next is \[\sqrt{2}\sin(\theta + \alpha)\] where \[\alpha=45\]

Answer 18

and r= root(2)

Answer 19

@sriramkumar.. how do we get \[\alpha=45\]?

Answer 20

because I've multiplied and divided each term by \[\sqrt{2}\] and took off the same as common. Now \[1/\sqrt{2} \] is sin45 and cos45, so we get \[\alpha = 45\]