# Solve Sinx +Cosx =1

No matter how I vị it, I always end up with \$x = 0, 90, 270\$ và \$360\$. All of those except \$270\$ is right, but I can"t quite figure out how to get the \$270\$ degrees out of the answer. I"ve tried using trig identities, I"ve tried squaring both sides, but I always kết thúc up with \$\$2sin xcos x\$\$ which then leads me to \$x = 0, 90, 270, 360\$. But \$\$sin (270) + cos (270) = -1\$\$ so I"m doing something wrong.

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Hint:

Use the formula \$sin(A+B) equiv sin A cos B + sin B cos A\$ lớn write \$sin x + cos x\$ in the size \$Rsin(x+alpha)\$, where \$R\$ & \$alpha\$ are numbers that you need to lớn find.

Once you have your \$R\$ và \$alpha\$, simply solve \$Rsin(x+alpha)=1\$.

The steps you followed are perfectly correct! The only thing you have to vị that you haven"t done already is kiểm tra for extraneous solutions. Whenever we square both sides, there is the chance that we get more solutions than we"re looking for.

So, you have correctly deduced that the menu \$0,90,270,360\$ (all angles in degrees) contains all potential solutions. After checking, you"ve noticed that \$270\$ is not a solution, but the rest are. So, the solutions are \$0,90,\$ and \$360\$.

As for why we get extra solutions: notice that although \$270\$ does not satisfy the original equation, we have\$\$(sin(270)+cos(270))^2=(-1)^2=1\$\$Which makes sense, since we just solved the equation \$\$(sin(x) + cos(x))^2 = 1\$\$

Another way to look at this is to lớn consider the equation as representing the intersection of two "curves" in polar coordinates, one being \$ r = 1 \$ (the unit circle), the other being the line \$ sin heta + cos heta = 1 Rightarrow r sin heta + r cos heta = r Rightarrow r = x + y , \$ with \$ r \$ set equal to 1 . The line has intercepts at ( 1 , 0 ) và ( 0 , 1 ) , meeting the circle at \$ heta = 0º extand heta = 90º \$ . (360º is considered khổng lồ be merely another "angle-name" for 0º , so it is not really a distinct solution.)

Use \$asin x + bcos x = sqrta^2+b^2cos(x- an^-1(frac ba))\$

\$sqrt2cos(x-fracpi4) = 1\$

\$cos(x-fracpi4) = frac1sqrt2\$

\$x - fracpi4 = 2npi pm fracpi4\$

Solving this gives 0,90,360 as solution.

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Hint. Square both sides lớn get \$\$(sin x+cos x)^2=1+2sin xcos x=1+sin 2x=1^2=1impliessin 2x=0.\$\$ The important thing here is to notice \$2sin xcos x=sin 2x\$. Notice I wrote an implication, not an equivalence, so we can get some extraneous solutions at the end. A routine check can spot all of them.

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Avoid squaring wherever practicable as it immediately introduces extraneous roots

Using Weierstrass substitution we have \$\$frac2t1+t^2+frac1-t^21+t^2=1\$\$ where \$t= anfrac x2\$

\$\$implies 2t+1-t^2=1+t^2iff t^2-t=0iff t=1,0\$\$

If \$displaystyle anfrac x2=0iff frac x2=n180^circiff x=n360^circ\$ where \$n\$ is any integer

If \$displaystyle anfrac x2=1iff frac x2=m180^circ+45^circiff x=m360^circ+90^circ\$ where \$m\$ is any integer

Using Double-Angle Formulas,

\$\$sin x+cos x=1implies 2sinfrac x2cosfrac x2=1-cos x=2sin^2frac x2\$\$

\$\$implies sinfrac x2left(cosfrac x2-sinfrac x2 ight)=0\$\$

\$\$(i)sinfrac x2=0implies frac x2=npi ext where n ext is any integer\$\$

\$\$(ii) cosfrac x2-sinfrac x2=0impliescosfrac x2=sinfrac x2iff anfrac x2=1\$\$ Find the rest in my other answer

\$sin x + cos x = 1 ightarrow (sin x + cos x )^2=1^2 iff colorgreenunderbrace(sin^2 x + cos^2 x)_=1+2sin x cos x =colorgreen1 iff sin x cos x = 0 \$

Now look at the unit circle khổng lồ see when this is true. After you have found the values of \$x\$ for which this holds, be sure to kiểm tra the outcome of \$sin x + cos x\$ for each of these values, to lớn eliminate the outcomes that generate value \$-1\$ (these values of \$x\$ came in because of the squaring - but are not truly solutions).

Expanding \$frac( extcis 75^circ- extcis 155^circ)(1-cos 8^circ+i sin 8 ^circ)2-2cos 8^circ\$
Trigonometric Equation Solve the equation \$( an heta +1)(sin^2 heta - sin heta) = 0\$ given that \$-pi leq θ leq 2pi\$
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