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Hey guys this is the question:

$$(\cos\theta - \sin\theta)^2 + (\cos\theta + \sin\theta)^2 = 2$$

I did this before but I forgot how I did it.. I tried to do this but don't know how to continue.. btw I can only use identity formulas, so please don't answer with other formulas I don't know about...

$$\cos^2\theta - 2\cos\theta\sin\theta + \sin^2\theta + 1 = 2$$

$$1 - 2\cos\theta\sin\theta = 2$$

then idk.. please help!

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    $\begingroup$ For a coder like you, it shouldn't be too hard to put your question in the appropriate format... $\endgroup$ – imranfat Apr 24 '16 at 20:18
  • $\begingroup$ Im Mainly a programmer @imranfat i dont know how to use the MathJax ... sorry.. $\endgroup$ – amanuel2 Apr 24 '16 at 20:18
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    $\begingroup$ Mathjax is much easier than programming.... $\endgroup$ – imranfat Apr 24 '16 at 20:30
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    $\begingroup$ Not wanting to learn??? $\endgroup$ – imranfat Apr 24 '16 at 20:50
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    $\begingroup$ Please read this tutorial on how to typeset mathematics using MathJax. Not writing your questions in MathJax is an imposition on the rest of us since it makes your questions difficult to read. $\endgroup$ – N. F. Taussig Apr 25 '16 at 9:35
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Set: $a=\cos \theta$, $b=\sin\theta$. Then:

$(a-b)^2 + (a+b)^2 = a^2 -2ab + b^2 + a^2 + 2ab + b^2 = 2a^2 + 2b^2 = 2(a^2 + b^2)=2(\sin\theta^2 + \cos\theta^2)=2.$

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hint: $(a-b)^2 = a^2 - 2ab+b^2, (a+b)^2 = a^2 + 2ab+b^2$, and $\sin^2 \theta + \cos^2 \theta = 1$. Use the above formulas with $a = \sin \theta, b = \cos \theta $.

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