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# Relations between Angles and sides of a triangle

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If in a triangle ABC, ab = c2  prove that

cos(A - B) + cos C + cos 2C = 1

NB - could only solve if a and b are equal to c [equilateal]...but not sure if correct reasoning.

Hope you are having a good day...regards

May 26, 2019
edited by OldTimer  May 26, 2019
edited by OldTimer  May 26, 2019

#1
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From the sine rule $$\displaystyle \frac{a}{\sin(A)}=\frac{b}{\sin(B)}=\frac{c}{\sin(C)},$$

we can say that

$$\displaystyle \frac{ab}{\sin(A)\sin(B)}=\frac{c^{2}}{\sin^{2}(C)},$$

and since we are told that $$\displaystyle ab=c^{2},$$

it follows that

$$\displaystyle \sin(A)\sin(B)=\sin^{2}(C) \dots(1)$$

Expanding the LHS of the (supposed) identity using standard trig identities, we have

$$\displaystyle \cos(A)\cos(B)+\sin(A)\sin(B)+\cos(C)+1-2\sin^{2}(C)$$

which equals, (making use of (1)),

$$\displaystyle \cos(A)\cos(B)-\sin(A)\sin(B)+\cos(C)+1$$

$$\displaystyle =\cos(A+B)+\cos(C)+1$$

$$\displaystyle =\cos(180-C)+\cos(C)+1$$

$$\displaystyle =-\cos(C)+\cos(C)+1$$

$$\displaystyle =1.$$

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May 27, 2019
#2
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Thanks that was great Tiggsy!!!

May 28, 2019