trig proof (1 Viewer)

Sylfiphy · Feb 11, 2024

Hey guys please help me prove this my test is in a few hours

show that the relationship sin(A+B) = sinA cosB+cosA sinB holds when A= 5π/3 and B=4π/3

Average Boreduser · Feb 11, 2024

literally just sub a and b in bruh

Sylfiphy · Feb 11, 2024

Average Boreduser said:
literally just sub a and b in bruh

if u haven’t realised already im dumb and when I sub it in I don’t get the same answer on LHS and RHS

Average Boreduser · Feb 11, 2024

Sylfiphy said:
if u haven’t realised already im dumb and when I sub it in I don’t get the same answer on LHS and RHS

I subbed them in and lhs = rhs r u in degrees per chance?

Sylfiphy · Feb 11, 2024

Average Boreduser said:
I subbed them in and lhs = rhs r u in degrees per chance?

u cant just say per chance and no I’m not in degrees pls show me what u did

Average Boreduser · Feb 11, 2024

I subbed a and b.

moonsuyoung · Feb 11, 2024

Show that the relationship sin(A+B) = sinA cosB+cosA sinB holds when A= 5π/3 and B=4π/3

Before subbing A and B, change them into degrees. One radian = 180 degrees.
Therefore, A = (5 * 180)/3 = 300 degrees and B = (4*180)/3 = 240 degrees

If you sub this in,
LHS = sin(A+B)
= sin(540)
= sin(180)
= 0

RHS = sin300cos240 + cos300sin240
= change to exact values
= 0

therefore, LHS = RHS

Luukas.2 · Feb 11, 2024

moonsuyoung said:
Show that the relationship sin(A+B) = sinA cosB+cosA sinB holds when A= 5π/3 and B=4π/3

Before subbing A and B, change them into degrees. One radian = 180 degrees.

Note that this should say that $\pi\ \text{radians} \equiv 180^{\circ}$ .

Also, it is important to get comfortable with working in radians as soon as you can. Calculus of trig functions does not work in degrees, and radians are not something added to irritate / confuse, they are essential for later work. One approach to this problem is to use the periodic nature of the sine and cosine functions to adjust any angle into the domain $x \in [0,\ 2\pi)$ . In other words, use that

$\begin{align*} \text{For any angle $\theta$:} \qquad \cos{\left(\theta - 2\pi\right)} &= \cos{\theta} \qquad \text{as the cosine function is periodic with period of $2\pi$ radians} \\ \sin{\left(\theta - 2\pi\right)} &= \sin{\theta} \qquad \text{as the sine function is periodic with period of $2\pi$ radians} \\ \tan{\left(\theta - \pi\right)} &= \tan{\theta} \qquad \text{as the tangent function is periodic with period of $\pi$ radians} \end{align*}$

to access angles which you are more familiar / comfortable:

$\begin{align*} \text{LHS} &= \sin{(A + B)} \\ &= \sin{\left(\frac{5\pi}{3} + \frac{4\pi}{3}\right)} \qquad \text{as we are given that $A = \frac{5\pi}{3}$ and that $B = \frac{4\pi}{3}$} \\ &= \sin{\left(\frac{9\pi}{3}\right)} \\ &= \sin{\left(3\pi\right)} \\ &= \sin{\left(3\pi - 2\pi\right)} \qquad \text{as $\sin{\theta}$ is a periodic function with period of $2\pi$} \\ &= \sin{\left(\pi\right)} \\ &= 0 \end{align*}$

Even within the domain $x \in [0,\ 2\pi)$ , trig functions of obtuse and reflex angles can be expressed in terms of acute angles using results like:

$\begin{align*} \text{For complementary angles:} \qquad \cos{\left(\frac{\pi}{2} - \theta\right)} &= \sin{\theta} \\ \sin{\left(\frac{\pi}{2} - \theta\right)} &= \cos{\theta} \\ \\ \text{For supplementary angles:} \qquad \cos{\left(\pi - \theta\right)} &= -\cos{\theta} \\ \sin{\left(\pi - \theta\right)} &= \sin{\theta} \\ \\ \text{Other angles:} \qquad \cos{\left(\pi + \theta\right)} &= -\cos{\theta} \\ \sin{\left(\pi + \theta\right)} &= -\sin{\theta} \\ \cos{\left(2\pi - \theta\right)} &= \cos{\theta} \\ \sin{\left(2\pi - \theta\right)} &= -\sin{\theta} \end{align*}$

Applying these in the given problem:

$\begin{align*} \text{RHS} &= \sin{(A + B)} \\ &= \sin{A}\cos{B} + \cos{A}\sin{B} \\ &= \sin{\frac{5\pi}{3}}\cos{\frac{4\pi}{3}} + \cos{\frac{5\pi}{3}}\sin{\frac{4\pi}{3}} \qquad \text{as we are given that $A = \frac{5\pi}{3}$ and that $B = \frac{4\pi}{3}$} \\ &= \sin{\left(\pi + \frac{2\pi}{3}\right)}\cos{\left(\pi + \frac{\pi}{3}\right)} + \cos{\left(\pi + \frac{2\pi}{3}\right)} \sin{\left(\pi + \frac{\pi}{3}\right)} \\ &= -\sin{\frac{2\pi}{3}} \times -\cos{\frac{\pi}{3}} + -\cos{\frac{2\pi}{3}}\times -\sin{\frac{\pi}{3}} \qquad \text{using $\sin{(\theta + \pi)} = -\sin{\theta}$ and $\cos{(\theta + \pi)} = -\cos{\theta}$} \\ &= \sin{\left(\pi - \frac{\pi}{3}\right)}\cos{\frac{\pi}{3}} + \cos{\left(\pi - \frac{\pi}{3}\right)}\sin{\frac{\pi}{3}} \\ &= \sin{\frac{\pi}{3}}\cos{\frac{\pi}{3}} - \cos{\frac{\pi}{3}}\sin{\frac{\pi}{3}} \qquad \text{using $\sin{(\pi - \theta)} = \sin{\theta}$ and $\cos{(\pi - \theta)} = -\cos{\theta}$} \\ &= 0 \\ &= \text{LHS} \qquad \text{as required} \end{align*}$

We could have used the actual values at the end, but it is not necessary as the final expression is clearly equal to zero.

Added... the results mentioned above for $\pi \pm \theta$ need not be memorised... knowing that trig functions are related in this kind of way, the results can be deduced from the graphs or by using transformations of functions from the syllabus.

WeiWeiMan · Feb 11, 2024

moonsuyoung said:
Show that the relationship sin(A+B) = sinA cosB+cosA sinB holds when A= 5π/3 and B=4π/3

Before subbing A and B, change them into degrees. One radian = 180 degrees.
Therefore, A = (5 * 180)/3 = 300 degrees and B = (4*180)/3 = 240 degrees

If you sub this in,
LHS = sin(A+B)
= sin(540)
= sin(180)
= 0

RHS = sin300cos240 + cos300sin240
= change to exact values
= 0

therefore, LHS = RHS

Converting to degrees is a kinda unnecessary step.
Also in the name of being pedantic, you should include the degrees sign when working with degrees.

trig proof (1 Viewer)

Sylfiphy

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Average Boreduser

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