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Trigonometric functions (both geometric and circular), relationships between lengths and angles in triangles and other topics relating to measuring triangles.

1 vote
3 answers
827 views

Deducing $\sin x + \cos x \ge 1$ from $(\sin x + \cos x)^2 = 1 + 2 \sin x \cos x$

So in trig, say I have an acute angle $X$. And one can intuitively conclude that $\sin x + \cos x \ge 1$, but how does the fact that $$(\sin x + \cos x)^2 = 1 + 2 \sin x \cos x$$ tell me that it is tr …
  • 1,361
0 votes
2 answers
4k views

What is the radius of a circle inscribed in a 6-8-10 triangle?

Here's me trying to do the problem. http://s10.postimg.org/bwwliium1/image.jpg So this problem was from a textbook and it was in the chapter of the theorem: If $\theta$ is the angle subtended by a c …
  • 1,361
1 vote
2 answers
248 views

Derive $\sin(\alpha - \beta)$

Here's me trying to do that. After $\frac{BP}{AC} = \frac{BP}{BC} \frac{BC}{AC} = \cos(\alpha - \beta)\tan(\beta)$, I didn't know what to do next. Theoretically, it should become $\cos(\alpha)\sin …
  • 1,361
2 votes
4 answers
2k views

Upon multiplying $\cos(20^\circ)\cos(40^\circ)\cos(80^\circ)$ by the sine of a certain angle...

So if $P = \cos(20^\circ)\cos(40^\circ)\cos(80^\circ)$, I can multiply $P$ by $\sin(X)$ so that the entire expression reduces to something manageable. I then take the simplified product and divide it …
  • 1,361
3 votes
1 answer
2k views

If $A + B + C = \pi$, then show that $\sin(A) + \sin(B) + \sin(C) = 4\cos\frac{A}{2}\cos\fra... [duplicate]

So i have $A + B + C = \pi$ $$\frac{A}{2} + \frac {B}{2} + \frac{C}{2} = \frac{\pi}{2}$$ $$4\cos\left(\frac{-B-C + \pi}{2}\right)\cos\left(\frac{-A -C + \pi}{2}\right)\cdots$$ And I doubt this lea …
  • 1,361
2 votes
5 answers
21k views

Prove that $\sin^2(A) - \sin^2(B) = \sin(A + B)\sin(A -B)$

This is my attempt: $$ \begin{align} & \phantom{={}}\sin^2(A) - \sin^2(B) = \sin(A + B)\sin(A -B) \\[8pt] & = (\sin(A)\cos(B)+\cos(A)\sin(B))(\sin(A)\cos(B) - \cos(A)\sin(B)) \\[8pt] & = (\sin(A)+\sin …
  • 1,361
0 votes
2 answers
100 views

Why is this derivation of $\int \sec x$ wrong?

I attempted to find the integral of $\sec x$ via u substitution. This is what I did: $$\int \sec x dx = \int \frac{1}{\cos x} \ dx$$ Let $u = \cos x$ $$du = -\sin x \ dx$$ $$-sinx \int \frac{1}{\cos …
  • 1,361