# If $9\sin\theta+40\cos\theta=41$ then prove that $41\cos\theta=40$.

I tried it this way:

$$40\cosθ+9\sinθ=41$$

$$9\sinθ=41-40\cos\theta$$

Squaring both the sides:

$$81\sin^2\theta=1681+1600\cos^2\theta-2\cdot 40\cdot 41 \cos\theta$$

$$81-81 \cos^2\theta= 1681+1600\cos^2\theta-3280 \cos\theta$$

$$1681\cos^2\theta-3280\cos\theta+1600=0$$

Solving the quadratic equation gives $\cos\theta=\dfrac{40}{41}$. It is not easy to solve the quadratic equation without calculator so there must be some other method, if yes then please explain.

P.S: I've found the other method so I am self-answering the question.

• So, You first typed answer then question? – user190024 Nov 6 '14 at 4:36
• @Infinity Actually simultaneously on satckedit.io – user103816 Nov 6 '14 at 4:37
• @user31782 There is an error in you quadratic. It should read $$1681\cos^2\theta-3280\cos\theta+1600=0$$. – John Joy Nov 7 '14 at 18:36

$$(9\sin\theta+40\cos\theta)^2+(40\sin\theta-9\cos\theta)^2=(40^2+9^2)(\sin^2\theta+\cos^2\theta)=41^2$$ as $41^2=40^2+9^2$

So, $40\sin\theta-9\cos\theta=0$ and $9\sin\theta+40\cos\theta=41$

Can you solve for $\cos\theta$

• Thankyou for the answer, it is quite interesting and of course short. $\cos\theta$ can be easily computed from $\tan\theta$. – user103816 Nov 6 '14 at 4:40
• $\tan\theta=9/40$ and then using $1+\tan^2\theta=sec^2\theta$. – user103816 Nov 6 '14 at 4:42
• @user31782, Squaring should generally be avoided as it immediately introduces extraneous root – lab bhattacharjee Nov 6 '14 at 4:44
• Do you mean that I should solve $40\sin θ−9\cos θ=0$ by using $9=r\sin y,40=r\cos y$? But we can use this substitution directly in $9\sin θ+40\cos θ$ as you did in the other answer. So what is the benefit of using the Brahmagupta-Fibonacci Idintity. – user103816 Nov 6 '14 at 9:17
• @user31782, Here we don't need to use $9=r\sin y$ etc. We just need to solve the two linear simultaneous equations for $\cos\theta$ – lab bhattacharjee Nov 6 '14 at 9:20

Squaring should generally be avoided as it immediately introduces extraneous root

Write $9=r\sin y,40=r\cos y$ where $r>0$

Squaring & adding we get, $r^2=41^2\implies r=41$

So, we have $r\cos(\theta-y)=41\iff\cos(\theta-y)=1=\cos0\implies\theta-y=2m\pi$ where $m$ is any integer

So, $\cos\theta=\cos(y+2m\pi)=\cos y=\dfrac{40}r$

Look at the work that you have already done. You know that $1681=41^2$, that $3280=2\cdot 40\cdot 41$, and that $1600=40^2$. You found this out when you squared $41-40\cos\theta$. It should now be quite easy to solve the quadratic $$1681\cos^2\theta−3280\cos\theta+1600=0$$ $$41^2\cos^2\theta−2(41)(40)\cos\theta−40^2=0$$ $$(41\cos\theta-40)^2=0$$ $$\dots$$

If you prefer to use the quadratic formula, remember to factor before you evaluate the square root.

$$\begin{array}{lll} \cos\theta &=& \frac{3280\pm\sqrt{3280^2-4(1681)(1600)}}{2\cdot 1681}\\ &=&\frac{2\cdot40\cdot41\pm\sqrt{(2\cdot40\cdot41)^2-4(41^2)(40^2)}}{2\cdot 41\cdot41}\\ &=&\frac{2\cdot40\cdot41}{2\cdot 41\cdot41}\\ &=&\frac{40}{41}\\ \end{array}$$

• I have done that. Actually when I solved the discriminant with calculator, I saw it becoming 0. I was solving 1681 sin^2 - 738 sin + 81 = 0 and then 9 sin = 41 - 40 cos. I squared the later one: $9^2 - 9^2\cos^2 x =41^2 +40^2 \cos^2 x - 2\cdot 41\cdot 40$ then $( 40^2 + 9^2) \cos^2 x- 2\cdot40\cdot41 + (41^2-9^2)$ Then I eventually noted $( 40^2 + 9^2)= 41^2$. Thankyou for your answer. – user103816 Nov 8 '14 at 7:33

I noted that $9,40$ and $41$ are pythagorean triple, i.e $9^2+40^2=41^2$. This fact can be used to solve the question easily. I'll show the solution for the general case:
$a\cos\theta+b\sin\theta=c,$ given that $a^2+b^2=c^2$

$$a\cos\theta+b\sin\theta=c$$ $$b\sin\theta=c-a\cos\theta$$ Squaring both the sides: $$b^2\sin^2\theta=c^2+a^2\cos^2\theta-2ca\cos\theta$$ Substituting $c^2$ with $a^2+b^2$ gives: $$b^2\sin^2\theta=a^2+b^2+a^2\cos^2\theta-2ca\cos\theta$$ $$a^2+b^2\cos^2\theta+a^2\cos^2\theta-2ca\cos\theta=0$$ $$a^2+(a^2+b^2)\cos^2\theta-2ca\cos\theta=0$$ $$a^2+(c\cos\theta)^2-2a(c\cos\theta)=0$$ $$(a-c\cos\theta)^2=0$$ $$\cos\theta=\dfrac{a}{c}$$

Substituting $c=41$ and $a=40$ will give the solution of the particular equation discussed in the question.

An interesting thing is that $\sin\theta=\sqrt{1-\dfrac{a^2}{c^2}}=\dfrac{b}{c},$ , i.e the equation actually belongs to a right angled triangle having $a$ as its base, $b$ as its prependicular and $c$ as its hypotenuse.

There is another method in which we divide both sides of the equation with $\sqrt{a^2+b^2}$ and then use the identity: $\sin (A+B)=\sin A \cos B+\cos A \cos B$

This method will give the same results.

Converse

Consider a right angled triangle having base $a$, perpendicular $b$ and hypotenuse $c$. We have:

$$\cos\theta=\dfrac{a}{c}\ \ \text{and}\ \ \ \sin\theta=\dfrac{b}{c}$$ $$\implies a=c\cos\theta \ \ \ \text{and}\ \ \ b=c\sin\theta$$

By the pythagoras theorem: $$a^2+b^2=c^2$$ $$\implies\ c^2\cos^2\theta+c^2\sin^2\theta=c^2$$ $$\implies c\cos\theta \cdot c\cos\theta + c\sin\theta \cdot c\sin\theta=c^2$$ $$\implies a\cos\theta+b\sin\theta=c$$

Since $41^2 - 40^2 = (41-40)(41+40) = 81 = 9^2$, there exists an angle $\phi$ satisfying $$\sin \phi = \frac{40}{41}, \quad \cos \phi = \frac{9}{41},$$ hence \begin{align*} 1 &= \frac{9}{41} \sin \theta + \frac{40}{41} \cos \theta \\ &= \sin\theta \cos\phi + \cos\theta \sin\phi \\ &= \sin(\phi+\theta),\end{align*} from which it follows that $\phi+\theta = \pi/2 + 2\pi k$ for some integer $k$, hence $$\cos\theta = \cos\left(\frac{\pi}{2} + 2\pi k - \phi\right) = \cos\left(\frac{\pi}{2} - \phi\right) = \sin\phi = \frac{40}{41},$$ and we are done.