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The equation of tangents to the curve `y=cos(x+y), -2pi |
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Answer» `x+2y=0` `m=-1/2` `y=cos(x+y)` differentiate with respect to x `dy/dx=-sin(x+y)(1+y^2)` `-1/2=-sin(x+y)(1-1/2)` `-1=-sin(x+y)` `sin(x+y)=1` `x+y=pi/2` `y=cospi/2=0` `sinx=1` `at x=-3/2pi,pi/2` `(-3/2pi,0)` and `(pi/2,0)` `y-0=-1/2(x+3/2pi)` `y=-x/2-3/4pi` `4y+2x+3pi=0` `y-0=-1/2(x-pi/2)` `y=-x/2+pi/4` `4y+2x-pi=0`. |
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