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# sin(25°)=cos(65°)

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1078
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sin(25°)=cos(65°)
tg(70°)=ctg(20°)
cos(35°)=sin(55°)
ctg(25°)=tg(65°)?

Guest Jan 12, 2015

#2
+92193
+5

Complementary Trig ratios

A really handy thing to know here is that the co in the front of half of the trig ratios stands for complement

Complementary angles add up to 90 degrees.

In any right angled triangle there are 2 acute angles.

If one is  θ    the other is the complement of  θ       That is (90-θ)

so

$$\\\mathbf{co}sine(\theta)=sin(90-\theta)\\ \mathbf{co}tangent(\theta)=tangent(90-\theta)\\ \mathbf{co}secant(\theta)=secant(90-\theta)\\$$

$$\\\mathbf{co}s(\theta)=sin(90-\theta)\\ \mathbf{co}t(\theta)=tan(90-\theta)\\ \mathbf{co}sec(\theta)=sec(90-\theta)\\$$

Melody  Jan 13, 2015
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#1
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yeah , this is the answer

Guest Jan 12, 2015
#2
+92193
+5

Complementary Trig ratios

A really handy thing to know here is that the co in the front of half of the trig ratios stands for complement

Complementary angles add up to 90 degrees.

In any right angled triangle there are 2 acute angles.

If one is  θ    the other is the complement of  θ       That is (90-θ)

so

$$\\\mathbf{co}sine(\theta)=sin(90-\theta)\\ \mathbf{co}tangent(\theta)=tangent(90-\theta)\\ \mathbf{co}secant(\theta)=secant(90-\theta)\\$$

$$\\\mathbf{co}s(\theta)=sin(90-\theta)\\ \mathbf{co}t(\theta)=tan(90-\theta)\\ \mathbf{co}sec(\theta)=sec(90-\theta)\\$$

Melody  Jan 13, 2015

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