Math Is Fun Forum

Geeky

Member

Registered: 2015-06-06

Posts: 9

Help in trigonometry

Well got some trigonometric task and need help in solving them....

If theta lies in Q3 then then cos theta/2 + sin theta/2 is +ve.. How to prove?

(sorry i am using old nokia 2700 so i can't use theta sign)

---

Math may be a bitter truth but truth is always useful.....

Bob

Administrator

Registered: 2010-06-20

Posts: 10,134

Re: Help in trigonometry

hi Geeky,

Welcome to the forum.

There is something missing from your first post.  Here is the graph of cos theta/2 + sin theta/2

As you can see, the function is another sine curve and is negative for half the time.

So what else does the question say?

Here is the proof that it is a sine curve:

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

Geeky

Member

Registered: 2015-06-06

Posts: 9

Re: Help in trigonometry

Thnx but i didn't undestand the step 2(how it came from step 1) of ur prove i mean to say that did u omit any step...

---

Math may be a bitter truth but truth is always useful.....

Bob

Administrator

Registered: 2010-06-20

Posts: 10,134

Re: Help in trigonometry

hi Geeky,

If you've met this technique before then all the steps are there.  But, as you are new to it, I'll give more details.

Have you met the formula:

If this is new to you,  I'll have to explain that as well, but let us just take it as correct for now.**

If you have an expression like this:

where p and q are numbers, you can always use this method to change the expression into a single sine expression.

What is needed is to make p into cos(B) and q into sin(B), but that is not likely to be possible.  They would have to be fractions for a start, and even if both p and q were fractions they would have to be the right fractions, so that inverse cos(p) and inverse sin(q) give the same angle B.

But there is a simple way to make p and q do what we want.

Calculate R = √ (p^2 + q^2) and then multiply and divide by R like this:

Because of the way that R was calculated it acts as the hypotenuse where p is the adjacent and q is the opposite, so this creates two fractions with the right property:

Because of the general properties of trig. functions this even works when p or q are negative. So we have

With your question p = q = 1, so that makes R = √2

Now sin(π/4) = cos(π/4) = 1/√2, so that's where step 2 comes from.

**If any of this is new to you, post back for an explanation.  It would help to have an idea about your current level of maths.

I am using radians ie. pi/4 rather than 45 degrees .

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

Geeky

Member

Registered: 2015-06-06

Posts: 9

Re: Help in trigonometry

I am known to other but not this one:

R{p/Rsin(A).q/Rcos(B)}=R{cosB.sinA+sinB.cosA}

---

Math may be a bitter truth but truth is always useful.....

Bob

Administrator

Registered: 2010-06-20

Posts: 10,134

Re: Help in trigonometry

Have a look at this:

https://www.youtube.com/watch?v=Tr8ck6qSXgc

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

Geeky

Member

Registered: 2015-06-06

Posts: 9

Re: Help in trigonometry

Thnx buddy

---

Math may be a bitter truth but truth is always useful.....