Calculus 1 problem?

Question

I need to find the equation of the tangent line to the curve $y=5sec(x)−10cos(x)y=5sec(x)−10cos(x)$ at the point $(π/3.5)$ . I need the answer in the form y=mx+b where m is the slope and b is the y-intercept. i've done other problems like these but for some reason i keep on getting this one wrong and i don't know why
thanks in advance

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Answer 1 · 2016-12-18T12:14:51

#y = 17.88x-14.26

graph{y cos x+ 5 cos 2x=0 [-10, 10, -5, 5]}

The constants are approximated to 4-sd.

$pi/3.5 radian = (180/3.5)^o=51.43^o$ , nearly

The given equation with x in radian is

$y=5sec x - 10 cos x=5(1-2cos^2x)/cos x=-5(cos 2x) / cos x.$

The point of contact of the tangent at $x = pi/3.5=0.8976$ is

$P( 0.8976, -5 (cos 103.86^o)/cos 51.43^o) = P(0.8976, 1.785)$

The slope m of the tangent at P is

y' = -5 (cos 2x / cox ) ' at P

$=-5((-2 cos x sin 2x+sin x cos 2x )/cos^2x)$ at P

$=-5(-2 cos 51.43^o sin 102.86^o + sin 51.43^o cos 102.86^0)/(cos^2 51.43^o)$

$=17.88$ . So, the equation of the tangent at P is

$y-1.785=17.88(x-0.8976)$ . Simplifying,

$y = 17.88x-14.26$