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 (IDENT)(FUNC) = (EQUIV)(FUNC)`

`\qquad \dfrac{IDENT}{FUNC} = \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 (EQUIV)(FUNC) = \left(EQUIV\right) \left(FUNC_SIMP\right)`

`\qquad \dfrac{EQUIV}{FUNC} = \dfrac{EQUIV}{FUNC_SIMP}`

This is `ANS`.
random() < 0.5 random() < 0.5 TAN ? ["\\tan^2\\theta + 1", "\\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 \tan^2\theta + 1 = \sec^2\theta`

`\qquad IDENT = EQUIV`

Plugging into our expression, we get

`\qquad (IDENT)(FUNC) = \left(EQUIV\right) \left(FUNC\right)`

`\qquad \dfrac{IDENT}{FUNC} = \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\right) = \left(EQUIV_SIMP\right) \left(FUNC_SIMP\right)`

`\qquad \dfrac{EQUIV}{FUNC} = \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\right) = \left(EQUIV_SIMP\right) \left(FUNC_SIMP\right)`

`\qquad \dfrac{EQUIV}{FUNC} = \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 (IDENT)(FUNC) = \left(EQUIV\right) \left(FUNC\right)`

`\qquad \dfrac{IDENT}{FUNC} = \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\right) = \left(EQUIV_SIMP\right) \left(FUNC_SIMP\right)`

`\qquad \dfrac{EQUIV}{FUNC} = \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\right) = \left(EQUIV_SIMP\right) \left(FUNC_SIMP\right)`

`\qquad \dfrac{EQUIV}{FUNC} = \dfrac{EQUIV_SIMP}{FUNC_SIMP}`

This is `ANS_SIMP = ANS`.
This is `ANS`.