Question

if acostheta-bsintheta=c then value of (asintheta+bcostheta) is
a.(a^2+b^2-c^2)^1/2
b.(a^2-b^2+c^2)^1/2
c..(a^2+b^2+c^2)^1/2
d..(b^2+c^2-a^2)^1/2

Answer 1

\[\begin{align*}a\sin\theta+b\cos\theta&=\sqrt{(a\sin\theta+b\cos\theta)^2}\\\\
&=\sqrt{a^2\sin^2\theta+2ab\sin\theta\cos\theta+b^2\cos^2\theta}\\\\
&=\sqrt{a^2\left(1-\cos^2\theta\right)+2ab\sin\theta\cos\theta+b^2\left(1-\sin^2\theta\right)}\\\\
&=\sqrt{a^2-a^2\cos^2\theta+2ab\sin\theta\cos\theta+b^2-b^2\sin^2\theta}\\\\
&=\sqrt{a^2+b^2-\left(a^2\cos^2\theta-2ab\sin\theta\cos\theta+b^2\sin^2\theta\right)}\\\\
&=\sqrt{a^2+b^2-\left(a\cos\theta-b\sin\theta\right)^2}\\\\
&=\sqrt{a^2+b^2-c^2}\\\\
\end{align*}\]

Answer 2

@Sandeepsra the porcess makes sense, I do not know haow to explain it any better!

Answer 3

oh and @SithsAndGiggles . nice job! couldn't have expalined it better myself, and nice usage of \(\LaTeX\) or the equation editor? idk but good job :)

Answer 4

Thanks @Jamierox4ev3r !

Answer 5

@Sandeepsra
\[\begin{align*}a\sin\theta+b\cos\theta&=\sqrt{(a\sin\theta+b\cos\theta)^2}\\\\
\end{align*}\]
This is like saying \(x=\sqrt{x^2}\). Make sense so far?

Answer 6

In the next step, I expand the binomial:
\[\begin{align*}(a\sin\theta+b\cos\theta)^2&=(a\sin\theta+b\cos\theta)(a\sin\theta+b\cos\theta)\\
&=a^2\sin^2\theta+2ab\sin\theta\cos\theta+b^2\cos^2\theta
\end{align*}\]

So our expression is the same as in the second line:
\[\sqrt{a^2\sin^2\theta+2ab\sin\theta\cos\theta+b^2\cos^2\theta}\]

Answer 7

Next, I use the Pythagorean identity, which says that \(\sin^2x=1-\cos^2x\) and \(\cos^2=1-\sin^2x\).

Doing so gives me
\[\sqrt{a^2\left(1-\cos^2\theta\right)+2ab\sin\theta\cos\theta+b^2\left(1-\sin^2\theta\right)}\]

Answer 8

Then I distribute:
\[a^2\left(1-\cos^2\theta\right)=a^2-a^2\cos^2\theta\]
I do the same for \(b^2\left(1-\sin^2\theta\right)\).

So I get
\[\sqrt{a^2-a^2\cos^2\theta+2ab\sin\theta\cos\theta+b^2-b^2\sin^2\theta}\]

Answer 9

The next step involves the commutative property of addition (\(a+b=b+a\)), where I simply move around some terms in the radicand:
\[\sqrt{a^2-a^2\cos^2\theta+2ab\sin\theta\cos\theta+b^2-b^2\sin^2\theta}\\
~~~~~~=\sqrt{a^2+b^2-a^2\cos^2\theta+2ab\sin\theta\cos\theta-b^2\sin^2\theta}\]
Next, I factor out a -1 from the last three terms.
\[\sqrt{a^2+b^2-\left(a^2\cos^2\theta-2ab\sin\theta\cos\theta+b^2\sin^2\theta\right) }\]

Answer 10

This next part is a lot like the first. The quantity within parentheses can be factored as a squared binomial:
\[\begin{align*}a^2\cos^2\theta-2ab\sin\theta\cos\theta+b^2\sin^2\theta&=(a\cos\theta-b\sin\theta)(a\cos\theta-b\sin\theta)\\
&=(a\cos\theta-b\sin\theta)^2\end{align*}\]
So I get
\[\sqrt{a^2+b^2-\left(a\cos\theta-b\sin\theta\right)^2}\]

Answer 11

Finally, I use the given expression for \(c\). Since \(c=a\cos\theta-b\sin\theta\), you get
\(\sqrt{a^2+b^2-c^2}\).

Answer 12

man if i had another medal i would give it to you, I am not even kidding @SithsAndGiggles , that was extremely through :P

Answer 13

Thanks again. Asker was a bit confused about the steps

Answer 14

haha apparently, although i don't see how you could know, he is not saying anything at all!

Answer 15

Sent some PMs