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# Pythagorean identities exercise

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Find all values of x in the interval [0,2pi) that satisfy the equation
sin^2(x)+cos^2(x)+tan^2(x)+cot^2(x)+sec^2(x)+csc^2(x)=7

Guest Feb 26, 2015

#2
+85789
+5

sin^2(x)+cos^2(x)+tan^2(x)+cot^2(x)+sec^2(x)+csc^2(x)=7

Note that  ...sin^2x + cos^2x = 1    ...so we have

tan^2x + cot^2x + sec^2x + csc^2x = 6

And

1 + tan^2x = sec^2x   ....so we have

2tan^2x + cot^2x + csc^2x = 5

And 1 + cot^2x = csc^x     so we have

2tan^2x + 2cot^x  = 4    divide through by 2

tan^2x + cot^2x = 2

So

sin^2x/ cos^2x + cos^2x/sin^2x = 2

[sin^4x + cos^4]  = 2sin^2xcos^2x    rearrange

sin^4x - 2sin^2xcos^x + cos^4x = 0   factor

(sin^2x -cos^2x)(sin^2x - cos^2x)  =0   setting one of the factors to 0, we have

sin^2x - cos^2x = 0   factor again

(sinx + cosx)(sinx-cosx)  = 0

For the first factor, we have sinx = -cosx.... and this occurs at 3pi/4 and at 7pi/4

For the second factor, we have  sinx = cos x  and this occurs at pi/4 and 5pi/4

So....our solutions are   pi/4, 3pi/4, 5pi/4 and 7pi/4

CPhill  Feb 26, 2015
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#1
+17721
+5

sin2(x) + cos2(x)  =  1

sec2(x)  =  1 + tan2(x)

csc2(x)  =  1 + cot2(x)

[sin2(x) + cos2(x)] + tan2(x) + cot2(x) + [sec2(x)] + [csc2(x)]  =  7

--->   1 + tan2(x) + cot2(x) + [1 + tan2(x)] + [1 + cot2(x)]  =  7

--->   3 + 2tan2(x) + 2cot2(x)  =  7

--->   2tan2(x) + 2cot2(x)  =  4

--->   tan2(x) + cot2(x)  =  2

--->   tan2(x) + 1 / tan2(x)  =  2

--->   tan4(x) + 1  =  2tan2(x)

--->   tan4(x) - 2tan2(x) + 1  =  0

--->   (tan2(x) - 1)(tan2(x) - 1)  =  0

--->   tan2(x)  =  1

--->   tan(x)  =  1    or    tan(x)  =  -1

--->   x  =  π/4  or  5π/4     or     x  =  3π/4  or  7π/4

geno3141  Feb 26, 2015
#2
+85789
+5

sin^2(x)+cos^2(x)+tan^2(x)+cot^2(x)+sec^2(x)+csc^2(x)=7

Note that  ...sin^2x + cos^2x = 1    ...so we have

tan^2x + cot^2x + sec^2x + csc^2x = 6

And

1 + tan^2x = sec^2x   ....so we have

2tan^2x + cot^2x + csc^2x = 5

And 1 + cot^2x = csc^x     so we have

2tan^2x + 2cot^x  = 4    divide through by 2

tan^2x + cot^2x = 2

So

sin^2x/ cos^2x + cos^2x/sin^2x = 2

[sin^4x + cos^4]  = 2sin^2xcos^2x    rearrange

sin^4x - 2sin^2xcos^x + cos^4x = 0   factor

(sin^2x -cos^2x)(sin^2x - cos^2x)  =0   setting one of the factors to 0, we have

sin^2x - cos^2x = 0   factor again

(sinx + cosx)(sinx-cosx)  = 0

For the first factor, we have sinx = -cosx.... and this occurs at 3pi/4 and at 7pi/4

For the second factor, we have  sinx = cos x  and this occurs at pi/4 and 5pi/4

So....our solutions are   pi/4, 3pi/4, 5pi/4 and 7pi/4

CPhill  Feb 26, 2015

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