# Trigonometry computingCalculate arctan5/12-arccos3/5

Asked on by shoppana

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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We'll put a = arctan 5/12 and b = arccos3/5.

So, we'll have to compute the difference between angles:

a - b

We'll apply the sine function to the difference above:

sin (a-b) = sin a*cos b - sin b*cos a

Since a = arctan 5/12, then tan a = 5/12.

Also, b = arccos3/5, then cos b = 3/5.

We'll apply the Pythagorean Theorem and we'll get:

(sin a)^2 + (cos a)^2 = 1

We'll divide by (cos a)^2 both sides and since tan a = sin a/cos a

(tan a)^2 + 1 = 1/(cos a)^2

(cos a)^2 = 1/[(tan a)^2 + 1]

(cos a)^2 = 1/(25/144 + 1)

(cos a)^2 = 144/(25+144)

(cos a)^2 = 144/169

cos a = +/- 12/13

Since tan a = sin a/cos a => sin a = tan a*cos a

sin a = (5/12)*(12/13)

sin a = +/-5/13

From the funcdamental formula of trigonometry, we'll calculate sin b.

cos b = 3/5

(sin b)^2 + (cos b)^2 = 1

(sin b)^2 = 1 - (cos b)^2

sin b = sqrt(1 - 9/25)

sin b = sqrt[(25-9)/25]

sin b =+/-4/5

Now, we can compute sin (a-b):

sin (a-b)=(5/13)*(3/5) - (4/5)(12/13)

sin (a-b) = (15 - 48)/5*13

sin (a-b) = -33/65

arctan 5/12 - arccos3/5 = -33/65