shuffle(["\\sin\\theta", "\\cos\\theta", "\\tan\\theta", "\\sec\\theta", "\\csc\\theta", "\\cot\\theta"]).slice(0,3) randFromArray(OPTIONS) random() < 0.5

(\sin^2 \theta + \cos^2 \theta)(FUNC) = \; ?
\dfrac{FUNC} {\sin^2 \theta + \cos^2 \theta} = \; ?

FUNC
  • op
We can use the identity \blue{\sin^2 \theta} + \orange{\cos^2 \theta} = 1 to simplify this expression.
init({ range: [[-1.2, 1.2], [-1.3, 1.3]], scale: 130 }); with(KhanUtil.currentGraph) { style({ stroke: "#ddd", strokeWidth: 1, arrows: "->" }); circle([0, 0], 1); line([-1.2, 0], [1.2, 0]); line([0, -1.2], [0, 1.2]); line([1.2, 0], [-1.2, 0]); line([0, 1.2], [0, -1.2]); style({ strokeWidth: 2.5, arrows: "" }); ang = 2*Math.PI/3; line([0, 0], [cos(ang), sin(ang)], {stroke: "black"}); label([cos(ang)/2, sin(ang)/2], "1", "above right"); line([0, 0], [cos(ang), 0], {stroke: ORANGE}); label([cos(ang), sin(ang)/2], "\\blue{\\sin\\theta}", "left"); line([cos(ang), 0], [cos(ang), sin(ang)], {stroke: BLUE}); label([cos(ang)/2, 0], "\\orange{\\cos\\theta}", "below"); arc([0,0], 0.2, 0, 120, {stroke: "black", arrows: "->"}); label([0,0.1], "\\theta", "above right"); }
We can see why this is true by using the Pythagorean Theorem.

So, (\sin^2 \theta + \cos^2 \theta)(FUNC) = 1 \cdot FUNC = FUNC

So, \dfrac{FUNC} {\sin^2 \theta + \cos^2 \theta} = \dfrac{FUNC}{1} = FUNC

random() < 0.5 random() < 0.5 SIN ? ["1 - \\sin^2\\theta", "\\cos^2\\theta"] : ["1 - \\cos^2\\theta", "\\sin^2\\theta"] trig.getOptionsResult(EQUIV, (MULT ? "*" : "/")) MULT ? trig.showSimplified(FUNC) : trig.showSimplified(FUNC, true)

(IDENT)(FUNC) = \; ?
\dfrac{IDENT}{FUNC} = \; ?

ANS
  • op
We can use the identity \blue{\sin^2 \theta} + \orange{\cos^2 \theta} = 1 to simplify this expression.
init({ range: [[-1.2, 1.2], [-1.3, 1.3]], scale: 130 }); with(KhanUtil.currentGraph) { style({ stroke: "#ddd", strokeWidth: 1, arrows: "->" }); circle([0, 0], 1); line([-1.2, 0], [1.2, 0]); line([0, -1.2], [0, 1.2]); line([1.2, 0], [-1.2, 0]); line([0, 1.2], [0, -1.2]); style({ strokeWidth: 2.5, arrows: "" }); ang = 2*Math.PI/3; line([0, 0], [cos(ang), sin(ang)], {stroke: "black"}); label([cos(ang)/2, sin(ang)/2], "1", "above right"); line([0, 0], [cos(ang), 0], {stroke: ORANGE}); label([cos(ang), sin(ang)/2], "\\blue{\\sin\\theta}", "left"); line([cos(ang), 0], [cos(ang), sin(ang)], {stroke: BLUE}); label([cos(ang)/2, 0], "\\orange{\\cos\\theta}", "below"); arc([0,0], 0.2, 0, 120, {stroke: "black", arrows: "->"}); label([0,0.1], "\\theta", "above right"); }
We can see why this is true by using the Pythagorean Theorem.

So, IDENT = EQUIV

Plugging into our expression, we get

\qquad (EQUIV)(FUNC)

\qquad \dfrac{EQUIV} {FUNC}

To make simplifying easier, let's put everything in terms of \sin and \cos. FUNC = FUNC_SIMP , so we can plug that in to get

\qquad \left(EQUIV\right) \left(FUNC_SIMP\right)

\qquad \dfrac{EQUIV} {FUNC_SIMP}

This is ANS.
random() < 0.5 random() < 0.5 TAN ? ["1 + \\tan^2\\theta", "\\sec^2\\theta"] : ["\\sec^2\\theta-1", "\\tan^2\\theta"] trig.getOptionsResult(EQUIV, (MULT ? "*" : "/")) [trig.showSimplified(FUNC, !MULT), trig.showSimplified(EQUIV, !MULT)] trig.showSimplified(ANS)

(IDENT)(FUNC) = \; ?
\dfrac{IDENT}{FUNC} = \; ?

ANS
  • op
We can derive a useful identity from \blue{\sin^2 \theta} + \orange{\cos^2 \theta} = 1 to simplify this expression.
init({ range: [[-1.2, 1.2], [-1.3, 1.3]], scale: 130 }); with(KhanUtil.currentGraph) { style({ stroke: "#ddd", strokeWidth: 1, arrows: "->" }); circle([0, 0], 1); line([-1.2, 0], [1.2, 0]); line([0, -1.2], [0, 1.2]); line([1.2, 0], [-1.2, 0]); line([0, 1.2], [0, -1.2]); style({ strokeWidth: 2.5, arrows: "" }); ang = 2*Math.PI/3; line([0, 0], [cos(ang), sin(ang)], {stroke: "black"}); label([cos(ang)/2, sin(ang)/2], "1", "above right"); line([0, 0], [cos(ang), 0], {stroke: ORANGE}); label([cos(ang), sin(ang)/2], "\\blue{\\sin\\theta}", "left"); line([cos(ang), 0], [cos(ang), sin(ang)], {stroke: BLUE}); label([cos(ang)/2, 0], "\\orange{\\cos\\theta}", "below"); arc([0,0], 0.2, 0, 120, {stroke: "black", arrows: "->"}); label([0,0.1], "\\theta", "above right"); }
We can see why this identity is true by using the Pythagorean Theorem.

Dividing both sides by \cos^2\theta, we get

\qquad \dfrac{\sin^2\theta}{\cos^2\theta} + \dfrac{\cos^2\theta}{\cos^2\theta} = \dfrac{1}{\cos^2\theta}

\qquad 1 + \tan^2\theta = \sec^2\theta

\qquad IDENT = EQUIV

Plugging into our expression, we get

\qquad \left(EQUIV\right) \left(FUNC\right)

\qquad \dfrac{EQUIV} {FUNC}

To make simplifying easier, let's put everything in terms of \sin and \cos. We know EQUIV = EQUIV_SIMP and FUNC = FUNC_SIMP , so we can substitute to get

\qquad \left(EQUIV\right) \left(FUNC_SIMP\right)

\qquad \left(EQUIV_SIMP\right) \left(FUNC_SIMP\right)

\qquad \dfrac{EQUIV} {FUNC_SIMP}

\qquad \dfrac{EQUIV_SIMP} {FUNC_SIMP}

To make simplifying easier, let's put everything in terms of \sin and \cos. We know EQUIV = EQUIV_SIMP, so we can substitute to get

\qquad \left(EQUIV\right) \left(FUNC_SIMP\right)

\qquad \left(EQUIV_SIMP\right) \left(FUNC_SIMP\right)

\qquad \dfrac{EQUIV} {FUNC_SIMP}

\qquad \dfrac{EQUIV_SIMP} {FUNC_SIMP}

This is ANS_SIMP = ANS.
This is ANS.
random() < 0.5 random() < 0.5 COT ? ["1 + \\cot^2\\theta", "\\csc^2\\theta"] : ["\\csc^2\\theta-1", "\\cot^2\\theta"] trig.getOptionsResult(EQUIV, (MULT ? "*" : "/")) [trig.showSimplified(FUNC, !MULT), trig.showSimplified(EQUIV, !MULT)] trig.showSimplified(ANS)

(IDENT)(FUNC) = \; ?
\dfrac{IDENT}{FUNC} = \; ?

ANS
  • op
We can derive a useful identity from \blue{\sin^2 \theta} + \orange{\cos^2 \theta} = 1 to simplify this expression.
init({ range: [[-1.2, 1.2], [-1.3, 1.3]], scale: 130 }); with(KhanUtil.currentGraph) { style({ stroke: "#ddd", strokeWidth: 1, arrows: "->" }); circle([0, 0], 1); line([-1.2, 0], [1.2, 0]); line([0, -1.2], [0, 1.2]); line([1.2, 0], [-1.2, 0]); line([0, 1.2], [0, -1.2]); style({ strokeWidth: 2.5, arrows: "" }); ang = 2*Math.PI/3; line([0, 0], [cos(ang), sin(ang)], {stroke: "black"}); label([cos(ang)/2, sin(ang)/2], "1", "above right"); line([0, 0], [cos(ang), 0], {stroke: ORANGE}); label([cos(ang), sin(ang)/2], "\\blue{\\sin\\theta}", "left"); line([cos(ang), 0], [cos(ang), sin(ang)], {stroke: BLUE}); label([cos(ang)/2, 0], "\\orange{\\cos\\theta}", "below"); arc([0,0], 0.2, 0, 120, {stroke: "black", arrows: "->"}); label([0,0.1], "\\theta", "above right"); }
We can see why this identity is true by using the Pythagorean Theorem.

Dividing both sides by \sin^2\theta, we get

\qquad \dfrac{\sin^2\theta}{\sin^2\theta} + \dfrac{\cos^2\theta}{\sin^2\theta} = \dfrac{1}{\sin^2\theta}

\qquad 1 + \cot^2\theta = \csc^2\theta

\qquad IDENT = EQUIV

Plugging into our expression, we get

\qquad \left(EQUIV\right) \left(FUNC\right)

\qquad \dfrac{EQUIV} {FUNC}

To make simplifying easier, let's put everything in terms of \sin and \cos. We know EQUIV = EQUIV_SIMP and FUNC = FUNC_SIMP , so we can substitute to get

\qquad \left(EQUIV\right) \left(FUNC_SIMP\right)

\qquad \left(EQUIV_SIMP\right) \left(FUNC_SIMP\right)

\qquad \dfrac{EQUIV} {FUNC_SIMP}

\qquad \dfrac{EQUIV_SIMP} {FUNC_SIMP}

To make simplifying easier, let's put everything in terms of \sin and \cos. We know EQUIV = EQUIV_SIMP, so we can substitute to get

\qquad \left(EQUIV\right) \left(FUNC_SIMP\right)

\qquad \left(EQUIV_SIMP\right) \left(FUNC_SIMP\right)

\qquad \dfrac{EQUIV} {FUNC_SIMP}

\qquad \dfrac{EQUIV_SIMP} {FUNC_SIMP}

This is ANS_SIMP = ANS.
This is ANS.