Express in terms of sine function of f(x)=sinx+cosx

[Elissa89]

f(x)=sinx+cosx

getting really frustrated with my math teacher. gives us forumlas for things but then barely shows us how to use them if at all and then throws problems at that we have to make sense of ourself. why can't math teachers teach?

anyway, the question is express f(x)=sinx+cosx in terms of sin.

 

[topsquark]

f(x)=sinx+cosx

getting really frustrated with my math teacher. gives us forumlas for things but then barely shows us how to use them if at all and then throws problems at that we have to make sense of ourself. why can't math teachers teach?

anyway, the question is express f(x)=sinx+cosx in terms of sin.

Probably what you are after is the relationship between sine and cosine. Using \(\displaystyle sin^2( \theta ) + cos^2( \theta ) =1\) then if \(\displaystyle \theta\) is in either Quadrant I or IV then \(\displaystyle cos( \theta ) = + \sqrt{1 - sin^2( \theta )}\), and if \(\displaystyle \theta\) is in Quadrant II or III then \(\displaystyle cos( \theta ) = - \sqrt{1 - sin^2( \theta )}\).

So \(\displaystyle sin( \theta ) + cos( \theta ) = sin( \theta ) \pm \sqrt{1 - sin^2( \theta )}\).

There is another relationship which I judge (from my own experience, anyway) that is not well known to students. I give it without proof (I don't know how to.)
\(\displaystyle a~sin( \theta ) + b~cos( \theta ) = c~sin( \theta + \phi )\)
where
\(\displaystyle c = \sqrt{a^2 + b^2}\) and \(\displaystyle \phi = atan2(b, a)\).

atan2 is similar to \(\displaystyle tan^{-1}\). It is defined here.

I'm guessing this is not what your instructor was after.

-Dan

 

[skeeter]

f(x)=sinx+cosx

anyway, the question is express f(x)=sinx+cosx in terms of sin.

$A\sin{x}+B\cos{x} = C\sin(x+\theta)$

$A\sin{x}+B\cos{x} = C\left[\sin{x}\cos{\theta}+\cos{x}\sin{\theta}\right]$

$A = C\cos{\theta}$, $B = C\sin{\theta}$ $\implies A^2+B^2 = C^2(\cos^2{\theta}+\sin^2{\theta}) \implies A^2+B^2=C^2(1) \implies C = \sqrt{A^2+B^2}$

$\dfrac{B}{A} = \dfrac{C\sin{\theta}}{C\cos{\theta}} = \tan{\theta} \implies \theta = \arctan\left(\dfrac{B}{A}\right)$

OK, you have $A=1$ and $B=1$ ... find the values of $C$ and $\theta$