Question

@Angle

Answer 1

\[\sin \theta + \cos \theta = 1\]

How would you solve this

Answer 2

Or maybe Sarah can use her brain for me

Answer 3

@umm ;)
I think she can.

Answer 4

No you

Answer 5

No u

Answer 6

Ms. Calculus

Answer 7

Uh
TGP IS CALLING ME GTG BYEEEE

Answer 8

\[0 \le \theta \le 2 \pi\]

Answer 9

solve for what ?

Answer 10

theta

Answer 11

Inky want to show me how to do this .-.

Answer 12

you want just the answer or work

Answer 13

use the identity \[\sin^2 \theta +\cos^2 \theta =1 \]

Answer 14

I want to understand, so the work would be good

Answer 15

wait that's a janky way to solve it

Answer 16

follow nnesha and then skip a few steps and use my answer

Answer 17

\[\large\rm \cos^2 \theta = \sin^2 \theta -1 , \rightarrow cos \theta = \sqrt{sin^2 \theta -1} \]
^substitute that
\[\large\rm \sin \theta +\cos \theta =1\]
\[\sin \theta + \sqrt {\sin^2 \theta -1}=1\]

Answer 18

\[\ \sqrt {\sin^2 \theta -1}=1-sin\theta \]
square both sides and then solve for sin theta.

Answer 19

ahh yeah that's the proper way to do it

Answer 20

granted every time you do that square root and square operation you have to justify it based on signs

Answer 21

\[\frac{ \pi }{ 2 }\] ?

Answer 22

ya

Answer 23

there is another solution. work??

Answer 24

oh yeah 0 i forgot

Answer 25

\[\frac{ 3\pi }{ 2 }\]

Answer 26

jk i'm a lazy liar and you should never trust me

Answer 27

2pi, not 3pi/2

Answer 28

Okay I guess I should write out my steps and you guys tell me where I screwed up, lol

Answer 29

The best way is to see the work in order to make the correction. otherwise ill be just guessing

Answer 30

\[\sin \theta + \cos \theta = 1\]
\[\sin \theta + \sqrt \sin^2 \theta - 1 = 1\]
Square it all
\[\sin ^2 \theta + \sin^2 \theta - 1 = 1\]
\[2 \sin^2 \theta = 2\]
\[\sin \theta = \pm 1\]

Referred to unit circle, and that's how I got

\[\frac{ \pi }{ 2 } , \frac{ 3 \pi }{ 2 }\]

Answer 31

Where I got +-1 from

\[2 \sin^2 \theta = 2 \]
\[\sin^2 \theta = 1\]
\[\sin^2 \theta - 1 = 0\]
\[(\sin \theta + 1)( \sin \theta - 1)\]
\[\sin \theta = -1, \sin \theta = 1\]

Answer 32

I'm kind of just going off copied notes for this since I was there for the lecture. Sorry lol. @Nnesha

Answer 33

wasnt'*

Answer 34

2nd step is wrong..

Answer 35

\(\color{#0cbb34}{\text{Originally Posted by}}\) @Nnesha
\[\large\rm \cos^2 \theta = \sin^2 \theta -1 , \rightarrow cos \theta = \sqrt{sin^2 \theta -1} \]
^substitute that
\[\large\rm \sin \theta +\cos \theta =1\]
\[\sin \theta + \sqrt {\sin^2 \theta -1}=1\]
\(\color{#0cbb34}{\text{End of Quote}}\)

we know that \[\cos^2 \theta + \sin^2 \theta =1 \]
solve for cos^2theta we will get \[\cos^2 \theta =1- \sin^2 \theta \]
to get rid of the square root from left side take square to the both side of the equation (including the *-1*)
\[\sqrt {\cos^2 \theta} =\sqrt{\sin^2 \theta -1}\]
\[\cos \theta = \sqrt{\sin^2 \theta -1}\]

Answer 36

Yeah I inputted that in for cos

Answer 37

\[\sin \theta + \sqrt \sin^2 \theta - 1 = 1\]
sqrt of sin^2theta -1 not just the sin^2theta

\[\sqrt{\sin^2 \theta-1} \cancel{=} \sin^2\theta\]

Answer 38

I know, but the sqrt doesn't extend all the way over for some reason.

Answer 39

ohh one second

Answer 40

I treated it as:

\[\sqrt (\sin^2 \theta - 1)\]

Answer 41

sorry that's right i was thinking about (sin^2theta -1)^2 *faceplm*

Answer 42

\[\sqrt {\sin^2\theta -1} = 1- \sin \theta \]
\[(\sqrt {\sin^2 \theta -1})^2 = (1-\sin \theta)^2\]
\[\sin^2 \theta -1= \]
(1-sintheta)^2 = ?

Answer 43

So the correct way to do this is to do the following steps:

Given:
\(\sin\theta + \cos\theta = 1\)

Square Both Sides:
\(\left(\sin\theta + \cos\theta\right)^2=1^2\)

Expand the above to get:
\(\sin^2\theta + \cos^2\theta + 2\sin\theta\cos\theta = 1\)

Remember that \(\sin^2\theta + \cos^2\theta = 1\) so:
\(1+ 2\cos\theta\sin\theta = 1\)

Subtract 1 from both sides:
\(2\sin\theta\cos\theta = 0\)

Divide both sides by 2:
\(\sin\theta\cos\theta = 0\)

Then use the product property (if ab = 0, then a = 0 or b = 0)

\(\sin\theta = 0\) or \(\cos\theta = 0\)

Answer 44

Now you should be able to easily find your theta.

Answer 45

ive heard of alpha, beta, omega, but no theta

Answer 46

In your third step, where does \[2 \sin \theta \cos \theta\] come from?

Answer 47

It comes from expanding the binomial. Remember algebra \((x + y)^2 = x^2 + 2xy + y^2\)

Answer 48

yeah, facepalm

Answer 49

wow

Answer 50

Thank you Hero, I get it now. Those steps were pretty easy to follow.

Answer 51

Don't forget to read that book I was telling you about.

Answer 52

Haha, I'll have time this summer to pour through books. Already have the great books you recommended on my reading list, I'll add that other one as well.

Answer 53

Ah, I see. Good luck.

Answer 54

That was an interesting question..\[\rm \sin^2 \theta +\cos ^2 \theta=1 \rightarrow cos \theta =\color{orange}{ \sqrt{1-sin^2 \theta}}\]\[\sin \theta + \color{orange}{\sqrt{1-\sin^2 \theta}}=1\]\[\rm \sqrt{1-\sin^2\theta}=1-\sin \theta\]\[\rm (\sqrt{1-\sin^2\theta})^2=(1-\sin \theta)^2\]\[\ \rm 1-\sin^2\theta = \sin^2\theta -2sin \theta +1\]\[2\sin \theta (\sin \theta -1)=0 \]\[\sin \theta = 0 ~,~ \sin \theta =1\]
This messy approach is *algebraically* correct...however it gives an extra solution which does not satisfy the original equation, an *extraneous solution*..

Answer 55

Except @Nnesha, your approach will not output possible values for theta that will satisfy the \(\cos(\theta)\) requirement. With my approach, we have

\(\theta = \sin^{-1}(0) = 0\)

\(\theta = \cos^{-1}(0) = \dfrac{\pi}{2}\)

And thus \(\sin(0) + \cos(0) = 1\)
or
\(\sin\left(\dfrac{\pi}{2}\right) + \cos\left(\dfrac{\pi}{2}\right) = 1 \)

So my approach finds two values of \(\theta\) that satisfy the equation.