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Special Sum of Trigonometric Functions

 
 
JinTina的头像
Special Sum of Trigonometric Functions
JinTina - 2024年01月23日 Tuesday 21:16
 

Hello,

In the class 9 of MC III Algebra, Dr. Wang said that:

-20≤sqrt(5)cos(a)+sqrt(15)sin(a)≤20

but why is this true? I tried "googling" a bit but didn't find any answers.

Additionally, can you help with 2.16, 2.28, 2.10, 3.13, 3.22, 5.20 5.23 5.27 5.28? These are all questions I asked before on the discussion forum and received a reply, but I am still stuck on it. I replied to each discussion post with what I'm stuck on and also what I already did.

Thanks!

Tina Jin

 
WangDr. Kevin的头像
Re: Special Sum of Trigonometric Functions
WangDr. Kevin - 2024年01月23日 Tuesday 23:24
 

Did I actually say that? :)  The general result related to this is:

$$-\sqrt{a^2+b^2}\leq a\cos\theta + b\sin\theta \leq \sqrt{a^2+b^2}.$$

To show this is true, draw a right triangle with $a$ and $b$ as its two legs of the right angle, and then the hypotenuse is $\sqrt{a^2+b^2}$.  Let $\phi$ be the angle opposite to $a$, then $\sin\phi =\dfrac{a}{\sqrt{a^2+b^2}}$ and $\cos\phi=\dfrac{b}{\sqrt{a^2+b^2}}$.  Thus

$$a\cos\theta + b\sin\theta = \sqrt{a^2+b^2}(\sin\phi\cos\theta+\cos\phi\sin\theta) = \sqrt{a^2+b^2}\sin(\phi+\theta).$$

So the inequality you showed above should have $-\sqrt{20}$ and $\sqrt{20}$ as bounds.  If I really said $-20$ and $20$, it was wrong (technically not wrong, just not helpful to solving the problem).

I am going to answer the other questions, just need to find the time.  Sorry about the delay.

JinTina的头像
Re: Special Sum of Trigonometric Functions
JinTina - 2024年01月24日 Wednesday 20:28
 

Oh, okay! I understand now. 

Great! Thank you, please take your time.