My Math Forum Simplying Trig

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 November 25th, 2011, 03:17 PM #1 Newbie   Joined: Feb 2011 Posts: 14 Thanks: 0 Simplying Trig (2cosx+5sinx)^2 + (5cosx-2sinx)^2 The answer has to be in numerical value This is what I have so far =2cos^2x + 2(2cosx5sinx) + 5sin^2x + 5cos^2x -2(5cosx2sinx) + 2sin^2x =2cos^2x + 4cosx10sinx + 5sin^2x + 5cos^2x -10cosx4sinx + 2sin^2x =7cos^2x + 7sin^2x + 4cosx10sinx - 10cosx4sinx Not sure what to do from here.
 November 25th, 2011, 03:25 PM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,923 Thanks: 1123 Math Focus: Elementary mathematics and beyond Re: Simplying Trig (2cos(x) + 5sin(x))^2 + (5cos(x) - 2sin(x))^2 = 4cos²(x) + 20cos(x)sin(x) + 25sin²(x) + 25cos²(x) - 20cos(x)sin(x) + 4sin²(x) Simplify and rearrange: = 4cos²(x) + 4sin²(x) + 25cos²(x) + 25sin²(x) = 4(cos²(x) + sin²(x)) + 25(cos²(x) + sin²(x)) apply Pythagorean identity, cos²(x) + sin²(x) = 1, = 4(1) + 25(1) = 29
 November 25th, 2011, 06:02 PM #3 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs Re: Simplying Trig Another way to proceed would be to use a linear combination identity: $$$a\cos x+b\sin x$$^2+$$b\cos x-a\sin x$$^2=$ $$$\sqrt{a^2+b^2}\sin\(x+\tan^{\small{-1}}\(\frac{a}{b}$$\)\)^2+$$\sqrt{b^2+(-a)^2}\sin\(x+\pi-\tan^{\small{-1}}\(\frac{b}{a}$$\)\)^2=$ $$$a^2+b^2$$$$\sin^2\(x+\tan^{\small{-1}}\(\frac{a}{b}$$\)+\sin^2$$x-\cot^{\small{-1}}\(\frac{a}{b}$$\)\)$ To simplify the second factor, let $c=\frac{a}{b}$: $\sin^2$$x+\tan^{\small{-1}}\(c$$\)+\sin^2$$x-\cot^{\small{-1}}\(c$$\)$ We know: $\tan^{\small{-1}}$$c$$+\cot^{\small{-1}}$$c$$=\frac{\pi}{2}$ thus: $-\cot^{\small{-1}}$$c$$=\tan^{\small{-1}}$$c$$-\frac{\pi}{2}$ giving us: $\sin^2$$x+\tan^{\small{-1}}\(c$$\)+\sin^2$$x+\tan^{\small{-1}}\(c$$-\frac{\pi}{2}\)$ Since $sin^2$$\theta-\frac{\pi}{2}$$=\cos^2$$\theta$$$ we may now write: $\sin^2$$x+\tan^{\small{-1}}\(c$$\)+\cos^2$$x+\tan^{\small{-1}}\(c$$\)=1$ and going back to the original problem, we now have: $$$a^2+b^2$$$$\sin^2\(x+\tan^{\small{-1}}\(\frac{2}{5}$$\)+\sin^2$$x-\cot^{\small{-1}}\(\frac{2}{5}$$\)\)=a^2+b^2$ In this problem, a = 2 and b = 5, thus it is evaluates to: $2^2+5^2=29$ The method used by greg1313 is by far more straightforward, I merely wanted to demonstrate an alternative...
 November 25th, 2011, 09:43 PM #4 Global Moderator   Joined: Dec 2006 Posts: 20,373 Thanks: 2010 There were two mistakes in the original work. The first was failing to square the coefficients 2 and 5 to get 4 and 25. The second was changing 2(2cosx5sinx) to 4cosx10sinx. Multiplication can be distributed over addition, but not over multiplication. The correct value would be 4cosx5sinx or 20cosxsinx. With these corrected, 29cos²x + 29sin²x could have been obtained, which is 29 (as cos²x + sin²x = 1).

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